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}262, numbers of green boxes, pink boxes, blue boxes and yellow boxes.
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Output
For each test case, output a line with the total number of boxes.
样例输入
4 1 2 3 4 0 0 0 0 1 0 0 0 111 222 333 404
样例输出
10 0 1 1070
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#define mod 998244353;
#define Max 0x3f3f3f3f;
#define Min 0xc0c0c0c0;
#define mst(a) memset(a,0,sizeof(a))
#define f(i,a,b) for(int i=a;i<b;i++)
using namespace std;
typedef long long ll;
const int maxn=100005;
ll temp;
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]='\0';
}
else
{
for (i=0;i<=max;i++)
c[i]=s[i];
c[i]='\0';
}
free(s);
}
int main(){
ios::sync_with_stdio(false);
int n;
while(cin>>n){
while(n--){
char a[105],b[105],c[105],d[105],ans[105];
cin>>a>>b>>c>>d;
add(a,b,ans);
add(ans,c,ans);
add(ans,d,ans);
cout<<ans<<endl;
}
}
return 0;
}