A + B Problem II |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 4112 Accepted Submission(s): 1434 |
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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.
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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.
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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. |
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Sample Input
2 1 2 112233445566778899 998877665544332211 |
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Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 |
今天很懵，一开始Java疯狂编译错误，WA，后来发现长度是1000，改用c来做；继续疯狂WA,这个题目这样都wa，我怕是傻子哟，最后发现一个参数弄错了。。。chiou；
除了中间的ac，其他都是我（微笑）
代码如下：
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char a[1005];
char b[1005];
int c[1005];
int main()
{
int t,lena,lenb,lenc;
cin>>t;
for(int i=1;i<=t;i++)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
int j;
scanf("%s%s",a,b);
lena=strlen(a)-1;
lenb=strlen(b)-1;
int index=0;
while(lena>=0&&lenb>=0)
{
c[index]=(c[index]+a[lena]+b[lenb]-'0'-'0');
c[index+1]=c[index]/10;
c[index]=c[index]%10;
lena--;
lenb--;
index++;
}
while(lena>=0)
{
c[index++]+=a[lena--]-'0';
}
while(lenb>=0)
{
c[index++]+=b[lenb--]-'0';
}
printf("Case %d:\n",i);
printf("%s + %s = ",a,b);
for(int l=index;l>=0;l--)
{
if(c[l]==0&&l==index)
continue;
printf("%d",c[l]);
}
printf("\n");
if(i<t)
printf("\n");
}
return 0;
}