A+B Problem II
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A+B Problem II
时间限制:3000 ms | 内存限制:65535 KB
难度:3
- 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.
A,B must be positive.
- 输入
- 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.
- 输出
- 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.
- 样例输入
21 2112233445566778899 998877665544332211
- 样例输出
Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
#include<stdio.h>#include<string.h>char a[1000],b[1000];int c[1000],d[1000],s[1000];int main(){ int t; scanf("%d",&t); int g=0; while(t--) { g++; memset(c,0,sizeof(c)); memset(d,0,sizeof(d)); memset(s,0,sizeof(s)); scanf("%s%s",a,b); int l,k; l=strlen(a); k=strlen(b); int p=0,m=0; for(int i=l-1;i>=0;i--) { c[p++]=a[i]-48; } for(int i=k-1;i>=0;i--) { d[m++]=b[i]-48; } int temp=0; int sum=0; int f=0; for(int i=0;i<1000;i++) { sum=c[i]+d[i]; if(sum>=10) { s[f++]=sum%10; temp=sum/10; c[i+1]=temp+c[i+1]; } else { s[f++]=sum; } } printf("Case %d:\n",g); printf("%s + %s = ",a,b); for(int i=f;i>=0;i--) { if(s[i]!=0) { for(int j=i;j>=0;j--) { printf("%d",s[j]); } printf("\n"); break; } } }}
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