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Problem C

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 3

Problem 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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
大数加法，主要意思是求两个数相加，这两个数可以达到32位，int可以达到21亿，__int64可以达到180亿都不能满足，所以必须得把这些数字用数组分开存储
比如a1a2a3--an与b1b2b3--bn相加将每一对应位相加，比如a1+b1判断有没有进位把进位加到a2+b2的和中，相同的a2+b2的进位加到a3+b3中，依次类推
#include<stdio.h>#include<iostream>#include<cstring>using namespace std;char a[1000],b[1000];int c[1000];int main(){int i,j,k,T,lena,lenb,temp,m,len;scanf("%d",&T);for(m=1;m<=T;m++){memset(c,0,sizeof(c));memset(a,'0',sizeof(a));memset(b,'0',sizeof(b));cin>>a>>b;lena=strlen(a);lenb=strlen(b);printf("Case ");printf("%d:\n",m);for(i=0;i<lena;i++)printf("%c",a[i]);printf(" + ");for(i=0;i<lenb;i++)printf("%c",b[i]);printf(" = ");for(i=lena-1,j=lenb-1,k=0;i>=0||j>=0;i--,j--,k++){if(i>=0&&j>=0)temp=a[i]+b[j]-'0'-'0';//这里a[]与b[]都为字符类型变为整形需要减去0的ascII码值，这样才能得整形结果else if(i>=0&&j<0)temp=a[i]-'0';//判断a[],b[]是否已经加完else if(j>=0&&i<0)temp=b[j]-'0';c[k]=temp+c[k];if(c[k]>=10){c[k+1]=c[k+1]+1;c[k]=c[k]-10;}}if(c[k]>=1)len=k;elselen=k-1;for(i=len;i>=0;i--)printf("%d",c[i]);if(m<T)printf("\n\n");elseprintf("\n");}return 0;}