Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 281441 Accepted Submission(s): 54185
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
#include<stdio.h>
#include<string.h>
#define N 1005
int common(int a, int b){
if(a<b){
return a;
}
else{
return b;
}
}
int rest(int a, int b){
int temp;
if(b>a){
temp = b;
b = a;
a = temp;
}
return (a-b-1);
}
int main()
{
char num1[N], num2[N], sum1[N], sum2[N]; // num存储输入的数,sum存储结果
int n;
while(scanf("%d", &n)!=EOF){ // n为要输入的行数
int k, i = 0, add=0;
for(; i<n; i++){
if(i){ // Output a blank line between two test cases.
printf("\n");
}
scanf("%s%s", num1, num2);
int len1 = strlen(num1); // 计算两个输入数的位数
int len2 = strlen(num2);
int long1 = len1-1, long2 = len2-1; // 对应的下标
if( len1 >= len2 ){
k = long2;
for(; k>=0; k--){
sum2[k] = (num1[long1]-""0"") + (num2[long2]-""0"")+add;
if(sum2[k]>=10){
sum2[k] -= 10;
add = ((num1[long1]-""0"" )+ (num2[long2]-""0"")+add)/10;
}
long1--;
long2--;
}
k = len1-len2-1;
for(; k>=0; k--){
sum1[k] = (num1[k]-""0"")+add;
if(sum1[k]>=10 && k>0){
sum1[k] -= 10;
add = ((num1[k]-""0"")+add)/10;
}
else{
add = 0;
}
}
}
else{
k = long1;
for(; k>=0; k--){
sum2[k] = (num1[long1]-""0"") + (num2[long2]-""0"")+add;
if(sum2[k]>=10){
sum2[k] -= 10;
add = ((num1[long1]-""0"" )+ (num2[long2]-""0"")+add)/10;
}
long1--;
long2--;
}
k = len2-len1-1;
for(; k>=0; k--){
sum1[k] = (num2[k]-""0"")+add;
if(sum1[k]>=10 && k>0){
sum1[k] -= 10;
add = ((num2[k]-""0"")+add)/10;
}
else{
add = 0;
}
}
}
int lenSum1 = rest(len1, len2);
int lenSum2 = common(len1, len2);
printf("Case %d:\n", i+1);
printf("%s + %s = ", num1, num2);
k = 0;
for(; k<=lenSum1; k++){
printf("%d", sum1[k]);
}
k = 0;
for(; k<lenSum2; k++){
printf("%d", sum2[k]);
}
printf("\n");
}
}
return 0;
}
