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=25;
const int mod=1e9+7;
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,0,sizeof(a));memset(b,0,sizeof(b));
        int i,j,k;
        for(i=len1-1,k=0;i!=-1;--i){
            a[k]=str1[i]-'0';
            k++;
        }
        for(j=len2-1,k=0;j!=-1;--j){
            b[k]=str2[j]-'0';
            k++;
        }
        for(i=0,up=0;i<25;++i){
            a[i]=a[i]+b[i]+up;
            up=a[i]/10;
            a[i]%=10;
        }
        memset(b,0,sizeof(b));memset(c,0,sizeof(c));
        for(i=len3-1,k=0;i!=-1;--i){
            b[k]=str3[i]-'0';
            k++;
        }
        for(j=len4-1,k=0;j!=-1;--j){
            c[k]=str4[j]-'0';
            k++;
        }
        for(int i=0,up=0;i<maxn;++i){
            b[i]=b[i]+c[i]+up;
            up=b[i]/10;
            b[i]%=10;
        }
        for(i=0,up=0;i<maxn;++i){
            a[i]=a[i]+b[i]+up;
            up=a[i]/10;
            a[i]%=10;
        }
        for(i=maxn-1;i>=0;i--){
            if(a[i])break;
        }
        if(i==-1)
            cout<<"0";
        else
            for(;i>=0;i--)
                cout<<a[i];
        cout<<endl;
    }
    return 0;
}