http://acm.hdu.edu.cn/showproblem.php?pid=1002
问题描述:
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
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.
问题分析:这道问题明显是两个大整数相加,我们可以直接利用高精度模板。鉴于这道题只考虑大数相加,于是我们可以简单利用两个数组分别表示大数。数组对应的下标表示不同的位数。
AC代码:
#include<iostream> using namespace std; #include <cmath> #include <cstring> #define N 1001 int main(){ int t; cin>>t; char c1[N],c2[N]; int a[N],b[N],c[N]; int len1,len2,len; for(int i = 1;i <= t;i++){ memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); cin>>c1>>c2; len1 = strlen(c1); len2 = strlen(c2); len = max(len1,len2); for(int j = len1 - 1;j >= 0;j--){ a[len1-1-j] = c1[j] - '0'; } for(int j = len2-1;j >= 0;j--){ b[len2-1-j] = c2[j] - '0'; } for(int j = 0; j < len;j++){ c[j] = a[j]+b[j]+c[j]; if(c[j] >= 10){ c[j] -= 10; c[j+1]++; } } if(c[len] == 0){ len--; } cout<<"Case "<<i<<":\n"<<c1<<" + "<<c2<<" = "; for(int j = len;j >= 0;j--){ cout<<c[j]; } if(i != t) cout<<endl<<endl; else cout<<endl; } return 0; }