2012/5/9 LEEREVWARE HDUACM
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Max Sum Plus Plus(1024)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9131 Accepted Submission(s): 2993
Problem Description
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.
Given a consecutive number sequence S1, S2, S3, S4 ... S x, ... S n(1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x≤ 32767). We define a function sum(i, j) = S i + ... + S j(1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(i m, j m) maximal (i x≤ i y≤ j x or i x≤ j y≤ j x is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... S n. Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1
2 3
2 6 -1 4 -2
3 -2 3
N!(1042)
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 31255 Accepted Submission(s): 8631
Problem Description
Given an integer N(0 ≤ N≤ 10000), your task is to calculate N!
Input
One N in one line, process to the end of file.
Output
For each N, output N! in one line.
Sample Input
1
2
3
Sample Output
1
2
6
#include
int main()
{
int i,j,n,u,arry;
while(scanf("%d",&n)!=EOF)
{
int a[10000]={1};
u=0;
for(i=1;i<=n;i++)
{
arry=0;
for(j=0;j<=u;j++)
{
arry+=a[j]*i;
a[j]=arry%10000;
arry/=10000;
}
if(arry)
{
u++;
a[u]=arry;
}
}
printf("%d",a[u]);
for(i=u-1;i>=0;i--)
printf("%04d",a[i]);
printf("\n");
}
return 0;
}
Max Sum(1003)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 75802 Accepted Submission(s): 17388
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4
Case 2:
7 1 6
#include
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
int i,j,t,n,m,x1,x,y,max,a;
scanf("%d",&t);
for(i=1;i<=t;i++)
{
scanf("%d",&n);
m=0;
x1=0;
x=0;
y=0;
max=-1000;
for(j=0;j { scanf("%d",&a); if(m<0) { m=a; x1=j; } else m+=a; if(max { max=m; x=x1; y=j; } } printf("Case %d:\n%d %d %d\n",i,max,x+1,y+1); } return 0; } Let the Balloon Rise(1004) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 40420 Accepted Submission(s): 14044 Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result. This year, they decide to leave this lovely job to you. Input Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters. A test case with N = 0 terminates the input and this test case is not to be processed. Output For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case. Sample Input 5 green red blue red red 3 pink orange pink Sample Output red pink #include #include int main() { //freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout); int i,j,k,n,l,max; char str[1000][15]; while(scanf("%d",&n)!=EOF&&n) { getchar(); int a[1000]={0}; char ch[1000][15]={""}; for(i=0;i scanf("%s",str[i]); strcpy(ch[0],str[0]); j=0; a[0]=1; for(i=1;i { l=0; for(k=0;k<=j;k++) if(strcmp(str[i],ch[k])==0) { l=1; a[k]++; break; } if(l==0) { j++; strcpy(ch[j],str[i]); a[j]++; } } max=0; for(i=0;i<=j;i++) if(max { max=a[i]; k=i; } printf("%s\n",ch[k]); } return 0; } Tick and Tick(1006) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4797 Accepted Submission(s): 1284 Problem Description The three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. Y ou are to calculate how much time in a day that all the hands are happy. Input The input contains many test cases. Each of them has a single line with a real number D between 0 and 120, inclusively. The input is terminated with a D of -1. Output For each D, print in a single line the percentage of time in a day that all of the hands are happy, accurate up to 3 decimal places. Sample Input 120 90 -1 Sample Output 100.000 0.000 6.251 严重超时（需改进算法） #include double g(double a,double b) { while(a>360) a-=360; while(b>360) b-=360; double min; if(a min=(b-a else min=(a-b return min; } double f(double a,double b,double c) { double min=g(a,b); min=(min min=(min return min; } int main() { //freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout); double t,s,m,h,a,num; double p; while(scanf("%lf",&a)!=EOF&&a!=-1) { num=0; for(t=0;t<43200;t+=0.01) { s=t*6; m=t/10.0; h=t/120.0; if(f(s,m,h)>=a) num+=0.01; } p=num/432.0; printf("%.3lf\n",p); } return 0; } 改进后 #include double g(double a,double b) { double min; if(a min=(b-a else min=(a-b return min; } double f(double a,double b,double c) { double min=g(a,b); min=(min min=(min return min; } int main() { //freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout); double t,s,m,h,a,num; double p; while(scanf("%lf",&a)!=EOF&&a!=-1) { num=0; s=0; m=0; h=0; for(t=0;t<43200;t+=0.1,s+=0.6,m+=0.01,h+=0.000833333) { if(s>360) s-=360; if(m>360) m-=360; if(f(s,m,h)>=a) num+=0.1; } p=num/432.0; printf("%.3lf\n",p); } return 0; } 钥匙计数之二(1480) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 692 Accepted Submission(s): 419 Problem Description 一把钥匙有N个槽，2 Input 本题无输入 Output 对2 Sample Output N=3: 104 N=4: 904 N=5: 5880 . . . . . N=25: 8310566473196300280 A + B Again(2057) Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8430 Accepted Submission(s): 3682 Problem Description There must be many A + B problems in our HDOJ , now a new one is coming. Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too. Easy ? AC it ! Input The input contains several test cases, please process to the end of the file. Each case consists of two hexadecimal integers A and B in a line seperated by a blank. The length of A and B is less than 15. Output For each test case,print the sum of A and B in hexadecimal in one line. Sample Input +A -A +1A 12 1A -9 -1A -12 1A -AA Sample Output 2C 11 -2C -90 A + B(1228) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7988 Accepted Submission(s): 4495 Problem Description 读入两个小于100的正整数A和B,计算A+B. 需要注意的是:A和B的每一位数字由对应的英文单词给出. Input 测试输入包含若干测试用例,每个测试用例占一行,格式为"A + B =",相邻两字符串有一个空格间隔.当A和B同时为0时输入结束,相应的结果不要输出. Output 对每个测试用例输出1行,即A+B的值. Sample Input one + two = three four + five six = zero seven + eight nine = zero + zero = Sample Output 3 90 96 #include #include int main() { int i,a,b; char ch[3],str1[2][10],str2[2][10],str[10][10]={"zero","one","two","three","four","five","six","seven","e ight","nine"}; while(1) { a=0; b=0; scanf("%s%s",str1[0],str1[1]); if(str1[1][0]=='+') { scanf("%s%s",str2[0],str2[1]); if(str2[1][0]=='=') for(i=0;i<10;i++) { if(strcmp(str[i],str1[0])==0) a=i; if(strcmp(str[i],str2[0])==0) b=i; } else { for(i=0;i<10;i++) { if(strcmp(str[i],str1[0])==0) a=i; if(strcmp(str[i],str2[0])==0) b+=i*10; if(strcmp(str[i],str2[1])==0) b+=i; } scanf("%s",ch); } } else { scanf("%s%s%s",ch,str2[0],str2[1]); if(str2[1][0]=='=') for(i=0;i<10;i++) { if(strcmp(str[i],str1[0])==0) a+=10*i; if(strcmp(str[i],str1[1])==0) a+=i; if(strcmp(str[i],str2[0])==0) b=i; } else { for(i=0;i<10;i++) { if(strcmp(str[i],str1[0])==0) a+=10*i; if(strcmp(str[i],str1[1])==0) a+=i; if(strcmp(str[i],str2[0])==0) b+=i*10; if(strcmp(str[i],str2[1])==0) b+=i; } scanf("%s",ch); } } if(a==0&&b==0) break; else printf("%d\n",a+b); } return 0; } Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 #include 杭州电子科技大学OJ题目分类 1001 整数求和水题 1002 C语言实验题——两个数比较水题 1003 1、2、3、4、5... 简单题 1004 渊子赛马排序+贪心的方法归并 1005 Hero In Maze 广度搜索 1006 Redraiment猜想数论：容斥定理 1007 童年生活二三事递推题 1008 University 简单hash 1009 目标柏林简单模拟题 1010 Rails 模拟题（堆栈） 1011 Box of Bricks 简单题 1012 u Calculate e 简单数学计算 1013 STAMPS 搜索or动态规划 1014 Border 模拟题 1015 Simple Arithmetics 高精度计算 1016 Shoot-out 博弈+状态压缩DP 1017 Tour Guide 1018 Card Trick 简单题 1019 Necklace Decomposition 贪心 1020 Crashing Robots 模拟题 1021 Electrical Outlets 简单题 1022 Watchdog 简单题 1023 Taxi Cab Scheme 图论：最小路径覆盖--->最大二分匹配1024 Pseudo-random Numbers 数论 1025 Card Game Cheater 简单题 1026 Investment 动态规划 1027 Pipes 1028 SETI 数学:高斯消元法 1029 Minimax Triangulation 计算几何 1030 Unequalled Consumption 母函数 1031 Declaration of Content 1032 Laserbox 搜索：DFS 1033 Bowlstack 1034 Pesky Heroes 1035 Reduced ID Numbers 暴力 1036 Tantrix 1037 Guardian of Decency 图论：匈牙利算法求二分图的最大匹配1038 Up the Stairs 简单数学题 1039 Sudoku 搜索：DFS 1040 The SetStack Computer 1041 Pie 二分法 1042 Ticket to Ride 动态规划 1043 The Bookcase 动态规划 1001 #include Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases. Sample Input 2 1 2 112233445566778899 998877665544332211 Sample Output Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 Author Ignatius.L #include 杭州电子科技大学OJ题目分类The Soul with Bone .: 1001 整数求和水题 1002 C语言实验题——两个数比较水题 1003 1、2、3、4、5... 简单题 1004 渊子赛马排序+贪心的方法归并 1005 Hero In Maze 广度搜索 1006 Redraiment猜想数论：容斥定理 1007 童年生活二三事递推题 1008 University 简单hash 1009 目标柏林简单模拟题 1010 Rails 模拟题（堆栈） 1011 Box of Bricks 简单题 1012 IMMEDIATE DECODABILITY Huffman编码 1013 STAMPS 搜索or动态规划 1014 Border 模拟题 1015 Simple Arithmetics 高精度计算 1016 Shoot-out 博弈+状态压缩DP 1017 Tour Guide 1018 Card Trick 简单题 1019 Necklace Decomposition 贪心 1020 Crashing Robots 模拟题 1021 Electrical Outlets 简单题 1022 Watchdog 简单题 1023 Taxi Cab Scheme 图论：最小路径覆盖--->最大二分匹配1024 Pseudo-random Numbers 数论 1025 Card Game Cheater 简单题 1026 Investment 动态规划 1027 Pipes 1028 SETI 数学:高斯消元法 1029 Minimax Triangulation 计算几何 1030 Unequalled Consumption 母函数 1031 Declaration of Content 1032 Laserbox 搜索：DFS 1033 Bowlstack 1034 Pesky Heroes 1035 Reduced ID Numbers 暴力 1036 Tantrix 1037 Guardian of Decency 图论：匈牙利算法求二分图的最大匹配1038 Up the Stairs 简单数学题 1039 Sudoku 搜索：DFS 1040 The SetStack Computer 1041 Pie 二分法 自己刷的题 这是我在杭电做题的记录，希望我的分享对你有帮助！！！ 1001 Sum Problem***********************************************************1 1089 A+B for Input-Output Practice (I)********************************2 1090 A+B for Input-Output Practice (II)********************************5 1091A+B for Input-Output Practice (III)****************************************7 1092A+B for Input-Output Practice (IV)********************************8 1093 A+B for Input-Output Practice (V)********************************10 1094 A+B for Input-Output Practice (VI)***************************************12 1095A+B for Input-Output Practice (VII)*******************************13 1096 A+B for Input-Output Practice (VIII)******************************15 How to Type***************************************************************16 1001 Sum Problem Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n. Input The input will consist of a series of integers n, one integer per line. Output For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer. 1111111杭电： 1000 A + B Problem (4) 1001 Sum Problem (5) 1002 A + B Problem II (6) 1005 Number Sequence (8) 1008 Elevator (9) 1009 FatMouse' Trade (11) 1021 Fibonacci Again (13) 1089 A+B for Input-Output Practice (I) (14) 1090 A+B for Input-Output Practice (II) (15) 1091 A+B for Input-Output Practice (III) (16) 1092 A+B for Input-Output Practice (IV) (17) 1093 A+B for Input-Output Practice (V) (18) 1094 A+B for Input-Output Practice (VI) (20) 1095 A+B for Input-Output Practice (VII) (21) 1096 A+B for Input-Output Practice (VIII) (22) 1176 免费馅饼 (23) 1204 糖果大战 (25) 1213 How Many Tables (26) 2000 ASCII码排序 (32) 2001 计算两点间的距离 (34) 2002 计算球体积 (35) 2003 求绝对值 (36) 2004 成绩转换 (37) 2005 第几天？ (38) 2006 求奇数的乘积 (40) 2007 平方和与立方和 (41) 2008 数值统计 (42) 2009 求数列的和 (43) 2010 水仙花数 (44) 2011 多项式求和 (46) 2012 素数判定 (47) 2014 青年歌手大奖赛_评委会打分 (49) 2015 偶数求和 (50) 2016 数据的交换输出 (52) 2017 字符串统计 (54) 2019 数列有序! (55) 2020 绝对值排序 (56) 2021 发工资咯：） (58) 2033 人见人爱A+B (59) 2037 今年暑假不AC (61) 2039 三角形 (63) 2040 亲和数 (64) 杭电ACM试题分类 枚举 1002 10041013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 10471048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 11061107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 11971200 1201 1202 1205 1209 1212（大数取模）1216 （链表）1218 1219 1225 1228 12291230 1234 1235 1236 1237 1239 1250 1256 1259 1262 1263 1265 1266 1276 1279 1282 1283 1287 1296 1302 1303 1304 1305 1306 1309 1311 1314 搜索，递归求解 1010 1016 1026 1043（双广）1044 （BFS+DFS） 1045 1067 1072 1104 1175 1180 1195 1208 1226 1238 1240 1241 1242 1258 1271 1312 1317 动态规划 1003 1024 1025 1028 1051 1058 1059 1069 1074 1078 1080 1081 1085 1087 1114 1158 1159 1160 1171 1176 1181 1203 1224 1227 1231 1244 1248 1253 1254 1283 1300 数学，递推，规律 1005 1006 1012 1014 1018 1019 1021 1023 1027 1030 1032 1038 1041 1046 1059 1060 1061 1065 1066 1071（微积分）1097 1098 1099 1100 1108 1110 1112 1124 1130 1131 1132 1134 1141 1143 1152 1155（物理题）1163 1165 1178 1194 1196（lowbit） 1210 1214 1200 1221 1223 1249 1261 1267 1273 1290 1291 1292 1294 1297 1313 1316 数论 1164 1211 1215 1222 1286 1299 计算几何 1086 1115 1147 ACM入门（杭电oj） Hdu 1000 #include a[j]=a[j]+str2[i]-'0'; a[j+1]=a[j+1]+a[j]/10; a[j]=a[j]%10; } count++; printf("Case %d:\n",count); printf("%s + %s = ",str1,str2); int flag=0; for(i=1004;i>=0;i--) if(flag||a[i]) { printf("%d",a[i]); flag=1; } printf("\n"); if(ca!=0) printf("\n"); } } Hdu 1003 #include 每个ACMer平时都是靠做题来训练的，现在就以杭州电子科技大学的OJ为例简单介绍一下如何使用OJ系统，在浏览器上输入杭电的网址https://www.doczj.com/doc/2f2950362.html, 首先，我们需要注册一个ID，点击Register new ID 填写好自己的个人信息之后，点击submit，就可以完成注册了，然后在主页右上角的登陆框输入ID，密码便可以登陆，这个时候点击主页上的 Problem Archive 打开OJ的题库 点击题号为 1000 的题目 A + B Problem Problem Description 一栏写的就是题目的描述，很多初学者一开始对全英文的题目会很不习惯，这是很正常的，多查查字典，很快就可以习惯了，而且也可以提高自己的英语阅读水平 Input 一栏写的是题目的输入要求，这里提示：每行将会有两个整数 A 和 B，输入进程一直延续到文件读完而结束（也就是要循环输入） Output 一栏写的是题目的输出要求，这里提示：对于每组数据，在一行中输入A+B的 值 接下来的 Sample Input 就是示例输入数据，而 Sample Output 就是相应的输出结果在读完题目之后，我们就可以在自己电脑上的编译器上编写代码，等编写好代码之后，点击页面中的Submit，接着把代码复制上去，再选择相应的语言 完成之后点击Submit，就可以看到OJ系统对我们的程序的判题结果了 这里返回了一个 Accepted，也就是说通过了OJ的判断，题目答对了！ 而OJ可能会返回的结果总共有： Accepted 通过 Wrong Answer 错误 Compilation Error 编译错误 Runtime Error 运行错误 Time Limit Exceeded 时间超出限制 Presentation Error 输出格式错误 Memory Limit Exceeded 内存超出限制 Output Limit Exceeded 输出了多余的东西 1001 Sum Problem (2) 1089 A+B for Input-Output Practice (I) (4) 1090 A+B for Input-Output Practice (II) (6) 1091 A+B for Input-Output Practice (III) (8) 1092 A+B for Input-Output Practice (IV) (9) 1093 A+B for Input-Output Practice (V) (11) 1094 A+B for Input-Output Practice (VI) (12) 1095 A+B for Input-Output Practice (VII) (13) 1096 A+B for Input-Output Practice (VIII) (14) 2000 ASCII码排序 (16) 2001计算两点间的距离 (17) 2002计算球体积 (19) 2003求绝对值 (20) 2004成绩转换 (21) 2005第几天 (22) 2006求奇数的乘积 (24) 2007平方和与立方和 (26) 2008数值统计 (27) 2009求数列的和 (28) 2010水仙花数 (29) 2011多项式求和 (31) 2012素数判定 (33) 2014青年歌手大奖赛_评委会打分 (34) 2015偶数求和 (36) 2016数据的交换输出 (38) 2017字符串统计 (40) 2019数列有序! (41) 2020绝对值排序 (43) 2021发工资咯：） (45) 2033人见人爱A+B (46) 2039三角形 (48) 2040亲和数 (49) 姓名：郑春杰 班级：电商1001 学号：34 【杭电ACM1000】 A + B Problem Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 # include Input The input will consist of a series of integers n, one integer per line. Output For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer. Sample Input 1 100 Sample Output 1 5050 # include ACM部分习题答案： A + B Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 100972 Accepted Submission(s): 33404 Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 # include 1000A + B Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 158161 Accepted Submission(s): 50186 Problem Description Calculate A + B. Input Each line will contain two integers A and B. Process to end of file. Output For each case, output A + B in one line. Sample Input 1 1 Sample Output 2 Author HDOJ Statistic | Submit | Discuss | Note #include 1002A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 84367 Accepted Submission(s): 15966 Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases. Sample Input 2 1 2 112233445566778899 998877665544332211 Sample Output Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 Author Ignatius.L Statistic | Submit | Discuss | Note 选修课考试作业 1001 Sum Problem ............................................................................................ 错误!未定义书签。1089 A+B for Input-Output Practice (I) .......................................................... 错误!未定义书签。1090 A+B for Input-Output Practice (II) ......................................................... 错误!未定义书签。1091 A+B for Input-Output Practice (III) ........................................................ 错误!未定义书签。1092 A+B for Input-Output Practice (IV) ........................................................... 错误!未定义书签。1093 A+B for Input-Output Practice (V) ......................................................... 错误!未定义书签。1094 A+B for Input-Output Practice (VI) ........................................................ 错误!未定义书签。1095 A+B for Input-Output Practice (VII) .......................................................... 错误!未定义书签。1096 A+B for Input-Output Practice (VIII) ...................................................... 错误!未定义书签。' 2000 ASCII码排序 ............................................................................................ 错误!未定义书签。2001计算两点间的距离.................................................................................. 错误!未定义书签。2002计算球体积 ............................................................................................. 错误!未定义书签。2003求绝对值 ................................................................................................. 错误!未定义书签。2004成绩转换 ................................................................................................. 错误!未定义书签。2005第几天 ..................................................................................................... 错误!未定义书签。2006求奇数的乘积 ......................................................................................... 错误!未定义书签。2007平方和与立方和...................................................................................... 错误!未定义书签。2008数值统计 ................................................................................................. 错误!未定义书签。2009求数列的和 ............................................................................................. 错误!未定义书签。~ 2010水仙花数 ................................................................................................. 错误!未定义书签。2011多项式求和 ............................................................................................. 错误!未定义书签。2012素数判定 ................................................................................................. 错误!未定义书签。2014青年歌手大奖赛_评委会打分................................................................ 错误!未定义书签。2015偶数求和 ................................................................................................. 错误!未定义书签。2016数据的交换输出...................................................................................... 错误!未定义书签。2017字符串统计 ............................................................................................. 错误!未定义书签。2019数列有序! ................................................................................................ 错误!未定义书签。2020绝对值排序.............................................................................................. 错误!未定义书签。2021发工资咯：）............................................................................................ 错误!未定义书签。: 2033人见人爱A+B .......................................................................................... 错误!未定义书签。2039三角形 ..................................................................................................... 错误!未定义书签。2040亲和数 ..................................................................................................... 错误!未定义书签。 姓名：郑春杰 班级：电商1001 hdu博弈，这些题都不难。 属于博弈简单题。 hdu1846巴什博弈，n%(m+1)==0先手必败。 #include #include(完整版)杭电acm部分答案
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