#include
#include
int main(){
char ax[20][1000], bx[20][1000], temp[1000], a[1000], b[1000];
int n, lena, lenb;
while(scanf("%d",&n)!= EOF){
for(int i = 0;i < n;i++){
scanf("%s %s",ax[i],bx[i]);
}
for(int i = 0;i < n;i++){
int num[1001] = {0};
strcpy(a,ax[i]);
strcpy(b,bx[i]);
lena = strlen(a);
lenb = strlen(b);
if(lena < lenb){
strcpy(temp,a);
strcpy(a,b);
strcpy(b,temp);
lena = strlen(a);
lenb = strlen(b);
}
int carry = 0;
int len = lena;
int add = 0;
for(int j = 0;j < len;j++){
if(lenb > 0){
add = a[lena-1] - '0' + b[lenb-1] - '0' + carry;
num[lena] = add % 10;
carry = add / 10;
lena--;
lenb--;
}
else{
add = a[lena-1] - '0' + carry;
num[lena] = add % 10;
carry = add / 10;
lena--;
lenb--;
}
}
num[0] = carry;
printf("case %d:\n",i+1);
printf("%s + %s = ",a,b);
if(num[0] == 0){
for(int j = 1;j <= len;j++){
printf("%d",num[j]);
}
printf("\n");
}
else{
for(int j = 0;j <= len;j++){
printf("%d",num[j]);
}
printf("\n");
}
printf("\n");
}
}
return 0;
}
经典的大数相加，题目如下：
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
程序跑图贴在下面，大牛们帮看看吧
