HDU 6225.Little Boxes-大数加法 (2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）)

整理代码。。。

Little Boxes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2304    Accepted Submission(s): 818

Problem Description

Little boxes on the hillside.
Little boxes made of ticky-tacky.
Little boxes.
Little boxes.
Little boxes all the same.
There are a green boxes, and b pink boxes.
And c blue boxes and d yellow boxes.
And they are all made out of ticky-tacky.
And they all look just the same.

 

Input

The input has several test cases. The first line contains the integer t (1 ≤ t ≤ 10) which is the total number of test cases.
For each test case, a line contains four non-negative integers a, b, c and d where a, b, c, d ≤ 2^62, indicating the numbers of green boxes, pink boxes, blue boxes and yellow boxes.

 

Output

For each test case, output a line with the total number of boxes.

 

Sample Input

4

1 2 3 4

0 0 0 0

1 0 0 0

111 222 333 404

 

Sample Output

10

0

1

1070

 

Source

2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

水题

代码:

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<cstdlib>
 #include<queue>
 #include<stack>
 #include<vector>
 using namespace std;
 typedef long long ll;
 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 const int maxn=;
 const int mod=1e9+;
 const int inf=0x3f3f3f3f;
 const double eps=acos(-1.0);
 int a[maxn],b[maxn],c[maxn],d[maxn];
 int main(int argc,const char *argv[]){
     string str1,str2,str3,str4;
     int len1,len2,len3,len4;
     int up,n;
     cin>>n;
     while(n--){
         cin>>str1>>str2>>str3>>str4;
         len1=str1.length();len2=str2.length();
         len3=str3.length();len4=str4.length();
         memset(a,,sizeof(a));memset(b,,sizeof(b));
         int i,j,k;
         for(i=len1-,k=;i!=-;--i){
             a[k]=str1[i]-'';
             k++;
         }
         for(j=len2-,k=;j!=-;--j){
             b[k]=str2[j]-'';
             k++;
         }
         for(i=,up=;i<;++i){
             a[i]=a[i]+b[i]+up;
             up=a[i]/;
             a[i]%=;
         }
         memset(b,,sizeof(b));memset(c,,sizeof(c));
         for(i=len3-,k=;i!=-;--i){
             b[k]=str3[i]-'';
             k++;
         }
         for(j=len4-,k=;j!=-;--j){
             c[k]=str4[j]-'';
             k++;
         }
         for(int i=,up=;i<maxn;++i){
             b[i]=b[i]+c[i]+up;
             up=b[i]/;
             b[i]%=;
         }
         for(i=,up=;i<maxn;++i){
             a[i]=a[i]+b[i]+up;
             up=a[i]/;
             a[i]%=;
         }
         for(i=maxn-;i>=;i--){
             if(a[i])break;
         }
         if(i==-)
             cout<<"";
         else
             for(;i>=;i--)
                 cout<<a[i];
         cout<<endl;
     }
     return ;
 }