A+B Problem II

時間限制：3000?ms ?|? 內存限制：65535?KB 難度：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.

輸入

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.

輸出

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.

樣例輸入

2 1 2 112233445566778899 998877665544332211

樣例輸出

Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

AC碼：

#include<stdio.h> #include<string.h> #include<stdlib.h> int main() {int T,i,k,len[2];char str[1010],ch[1010];int num[2][1010],max,count=0;scanf("%d",&T);while(T--){scanf("%s",str);scanf("%s",ch);len[0]=strlen(str);len[1]=strlen(ch);max=len[0]>len[1]?len[0]:len[1];memset(num,0,sizeof(num));for(i=0;str[i]!='\0';i++){num[0][len[0]]=str[i]-'0';len[0]--;}for(i=0;ch[i]!='\0';i++){num[1][len[1]]=ch[i]-'0';len[1]--;}for(i=1;i<=max;i++){num[0][i]=num[0][i]+num[1][i];if(num[0][i]>=10){num[0][i+1]+=1;num[0][i]=num[0][i]%10;}}printf("Case %d:\n",++count);printf("%s + %s = ",str,ch);if(num[0][max+1]!=0)printf("%d",num[0][max+1]);for(i=max;i>0;i--)printf("%d",num[0][i]);printf("\n");}return 0; }

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