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
题意：
给了四个long long数，求四个数之和。
题解：
long long范围内的数相加会溢出，所以用大整数加法，套模板就好了。
/*大数加法*/
# include<cstdio>
# include<cstring>
# include<cstdlib>
#include<iostream>
using namespace std;
void add(char* a,char* b,char* c)
{
int i,j,k,max,min,temp;
char *s,*pmax,*pmin;
max=strlen(a);
min=strlen(b);
if (max<min)
{
temp=max;
max=min;
min=temp;
pmax=b;
pmin=a;
}
else
{
pmax=a;
pmin=b;
}
s=(char*)malloc(sizeof(char)*(max+1));
s[0]='0';
for (i=min-1,j=max-1,k=max;i>=0;i--,j--,k--)
s[k]=pmin[i]-'0'+pmax[j];
for (;j>=0;j--,k--)
s[k]=pmax[j];
for (i=max;i>=0;i--)
if (s[i]>'9')
{
s[i]-=10;
s[i-1]++;
}
if (s[0]=='0')
{
for (i=0;i<=max;i++)
c[i-1]=s[i];
c[i-1]=' ';
}
else
{
for (i=0;i<=max;i++)
c[i]=s[i];
c[i]=' ';
}
free(s);
}
int main()
{
int n;
cin>>n;
while(n--)
{
char a[100],b[100],c[100],d[100],ans[100];
cin>>a>>b>>c>>d;
add(a,b,ans);
add(ans,c,ans);
add(ans,d,ans);
cout<<ans<<endl;
}
return 0;
}