大数相加
发布时间:2020-12-14 03:03:09 所属栏目:大数据 来源:网络整理
导读:原题http://acm.hdu.edu.cn/showproblem.php?pid=1047 Integer Inquiry Time Limit: 2000/1000 MS (Java/Others)??? Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12692??? Accepted Submission(s): 3214 Problem Description One of th
原题http://acm.hdu.edu.cn/showproblem.php?pid=1047 Integer InquiryTime Limit: 2000/1000 MS (Java/Others)??? Memory Limit: 65536/32768 K (Java/Others)
Problem Description
One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment,once one became available on the third floor of the Lemon Sky apartments on Third Street.)
Input
The input will consist of at most 100 lines of text,each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length,and will only contain digits (no VeryLongInteger will be negative).
The final input line will contain a single zero on a line by itself.
Output
Your program should output the sum of the VeryLongIntegers given in the input.
This problem contains multiple test cases! The first line of a multiple input is an integer N,then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks. The output format consists of N output blocks. There is a blank line between output blocks.
Sample Input
Sample Output
#include <stdio.h> #include <stdlib.h> #include <limits.h> #include <malloc.h> #include <limits.h> #include <ctype.h> #include <string.h> #include <string> #include <algorithm> #include <iostream> #include <queue> #include <deque> #include <stack> #include <vector> #include <set> #include <map> using namespace std; #define MAXN 500 + 10 char a[MAXN],b[MAXN],c[MAXN],sum[MAXN]; //int count; int mark; void jia(char sum[],char a[]){ int i,j,z,k,count; k = 0; z = 0; for(i=strlen(sum)-1,j=strlen(a)-1;i>=0||j>=0;i--,j--){ if(i >= 0) z+=sum[i]-'0'; if(j >= 0) z+=a[j]-'0'; c[k++] = z%10+'0'; z = z/10; } if(z){ c[k++] = '1'; } c[k] = ' '; i = 0; count = k; // count = 0; for(--k;k>=0;k--){ sum[i++] = c[k]; // count++; } sum[i] = ' '; if(mark == 0){ for(i=0;i<count;i++){ if(i != count-1){ printf("%c",sum[i]); } else{ printf("%cn",sum[i]); } } } } int main(){ int T,i,cas; //int mark; while(~scanf("%d",&T)){ for(cas=1;cas<=T;cas++){ mark = 1; memset(a,sizeof(a)); memset(b,sizeof(b)); memset(c,sizeof(c)); memset(sum,sizeof(sum)); if(cas > 1){ printf("n"); } while(mark != 0){ scanf("%s",a); if(strcmp(a,"0") == 0){ mark = 0; jia(sum,a); } else{ jia(sum,a); } } } } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |