又一版 A+B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14701    Accepted Submission(s): 5618 Problem Description 输入两个不超过整型定义的非负10进制整数A和B(<=231-1),输出A+B的m (1 < m <10)进制数.   Input 输入格式:测试输入包含…

Problem Description 输入两个不超过整型定义的非负10进制整数A和B(<=231-1),输出A+B的m (1 < m <10)进制数. Input 输入格式:测试输入包含若干测试用例.每个测试用例占一行,给出m和A,B的值. 当m为0时输入结束. Output 输出格式:每个测试用例的输出占一行,输出A+B的m进制数. Sample Input 8 1300 48 2 1 7 0 Sample Output 2504 1000 java强大的Integer.toStri…

另一个版本 A+B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12894    Accepted Submission(s): 4901 Problem Description 输入两个不超过整型定义的非负10进制整数A和B(<=231-1).输出A+B的m (1 < m <10)进制数.   Input 输入格式:測试输…

题目 一个进制转换的题,注意0+0的情况 代码如下: #include <cstdio> int d[1000]; void solve(int n,int base) { int p = 0; while(n) { d[p++]=n%base; n=n/base; } for(int i=p-1;i>=0;i--) printf("%d",d[i]); printf("\n"); } int main() { int m,a,b; while(~…

看了http://lovnet.iteye.com/blog/1690276的答案 好巧妙的方法 递归实现十进制向m进制转换 #include "stdio.h" int m; void CK(int n) { if(n>=m) CK(n/m); printf("%d",n%m); } int main() { int a,b; while(~scanf("%d",&m)&&m) { scanf("%d%d…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…

转载来自:http://www.cppblog.com/acronix/archive/2010/09/24/127536.aspx 分类一: 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029.1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093.1094.1095.1096.1097.1098.1106.1108.1157…

模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201 120…

King's Game 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5643 Description In order to remember history, King plans to play losephus problem in the parade gap.He calls n(1≤n≤5000) soldiers, counterclockwise in a circle, in label 1,2,3...n. The firs…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…