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HDU 1002 A + B Problem II

2016-03-10 18:32 411 查看

A + B Problem II

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 16104 Accepted: 4547

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

思路：

把数值每一位拆开存进数组，利用ASCII码转换数据，然后进行相加。

#include<stdio.h>
#include<string.h>
int main()
{
char a[1003];
char b[1003];
char ans[1003];
int T;
scanf("%d",&T);
int cas = 0 ;
while(T--)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(ans,0,sizeof(ans));
scanf("%s",a);
scanf("%s",b);
strrev(a);
strrev(b);
int i;
for(i = 0 ; a[i]&&b[i];i++)
{
ans[i]+=a[i]+b[i]-48;
if(ans[i]>'9'){
ans[i+1]++;
ans[i]-=10;
}
}
while(a[i]){
ans[i]+=a[i];
if(ans[i]>'9'){
ans[i+1]++;
ans[i]-=10;
}
i++;
}
while(b[i]){
ans[i]+=b[i];
if(ans[i]>'9'){
ans[i+1]++;
ans[i]-=10;
}
i++;
}
if(ans[i]==1)ans[i]+=48;
strrev(ans);
strrev(a);
strrev(b);
if(cas)printf("\n");
printf("Case %d:\n",++cas);
printf("%s + %s = %s\n",a,b,ans);
}
}

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