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Time Limit:1000MS
Memory Limit:32768KB
Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
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.
Output
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. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
以下是代码：
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#include <iostream>#include <cstring>#include <cstdio>using namespace std;char a[1000],b[1000];int c[1100],n;void cal(char a[],char b[],int r){ int i = strlen(a),j = strlen(b); int k=0,t; i--;j--; memset(c,0,sizeof(c)); while(i>=0 || j >=0){ if(i>=0 && j>=0)t=a[i]+b[j]+c[k]-‘0‘-‘0‘; else if (i<0)t=b[j]+c[k]-‘0‘; else t=a[i]+c[k]-‘0‘; if(t>=10){ c[k++]=t%10;c[k]+=1; }else c[k++]=t; i--;j--; } while(c[k]==0)k--; printf("Case %d:\n%s + %s = ",r,a,b); for(i=k;i>=0;i--)printf("%d",c[i]); printf("\n"); if(r!=n)printf("\n");}int main(){ int len1,len2; cin >> n; for(int i=0;i<n;i++){ scanf("%s%s",a,b); cal(a,b,i+1); }} |
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原文地址：http://www.cnblogs.com/gzdaijie/p/4298088.html
