A + B Problem II

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
2
1 2
112233445566778899 998877665544332211

Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

大数问题，用字符串模拟即可

#include"stdio.h"#include"string.h"int main(){    char a[10001],b[10001];    int c[10001],j,k,g,h;    int n,Al,Bl,sum;    scanf("%d",&n);    for (int i=1;i<=n;i++)    {        scanf("%s %s",a,b);        getchar();        Al=strlen(a);        Bl=strlen(b);        if(i!=1)        printf("\n");        printf("Case %d:\n",i);        printf("%s",a);        printf(" + ");        printf("%s",b);        printf(" = ");        if (Al>=Bl)        {            for (j=Bl-1;j>=0;j--)            {                b[j+Al-Bl]=b[j];            }            for (k=0;k<Al-Bl;k++)            {                b[k]='0';            }            for (g=Al-1;g>0;g--)            {                sum=a[g]+b[g]-96;                if (sum>=10)                {                    c[g]=sum-10;                    a[g-1]=a[g-1]+1;                }                else                    c[g]=sum;            }            if (a[0]+b[0]-96>=10)            {                c[0]=a[0]+b[0]-106;                printf("1");                for (h=0;h<Al;h++)                {                    printf("%d",c[h]);                }            }            else            {                c[0]=a[0]+b[0]-96;                for (h=0;h<Al;h++)                {                    printf("%d",c[h]);                }            }            printf("\n");        }        else        {            for (j=Al-1;j>=0;j--)            {                a[j+Bl-Al]=a[j];            }            for (k=0;k<Bl-Al;k++)            {                a[k]='0';            }            for (g=Bl-1;g>0;g--)            {                sum=a[g]+b[g]-96;                if (sum>=10)                {                    c[g]=sum-10;                    b[g-1]=b[g-1]+1;                }                else                    c[g]=sum;            }            if (a[0]+b[0]-96>=10)            {                c[0]=a[0]+b[0]-106;                printf("1");                for (h=0;h<Al;h++)                {                    printf("%d",c[h]);                }            }            else            {                c[0]=a[0]+b[0]-96;                for (h=0;h<Bl;h++)                {                    printf("%d",c[h]);                }            }            printf("\n");        }          }    return 0;}