版权声明:仅供研究,转载请注明出处。 https://blog.csdn.net/CSUstudent007/article/details/82147921
思路:
将前两个数组的数加起来(ab),后两个数组数加起来(cd),然后枚举ab的值,二分cd的数进行判断。
ps:如果cd找到与ab相符合的数
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[4001],b[4001],c[4001],d[4001];
int ab[4000*4000+1],cd[4000*4000+1];
int k2;
int check(int x)
{
int left=1,right=k2-1,mid;
while (left<=right)
{
mid=(left+right)/2;
if (x==cd[mid])
{
int w=0,e=mid;
while (x==cd[e]&&e<k2)
e++,w++;
e=mid-1;
while (x==cd[e]&&e>0)
e--,w++;
return w;
}
else if (x<cd[mid])
right=mid-1;
else
left=mid+1;
}
return 0;
}
int main()
{
int t,i,j,q;
while (~scanf("%d",&t))
{
for (i=1;i<=t;i++)
scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
memset(ab,0,sizeof(ab));
memset(cd,0,sizeof(cd));
int k1=1,sum=0;
k2=1;
for (i=1;i<=t;i++)
{
for (j=1;j<=t;j++)
{
ab[k1++]=a[i]+b[j];
cd[k2++]=-(c[i]+d[j]);
}
}
sort(cd+1,cd+k2);
for (i=1;i<k1;i++)
sum+=check(ab[i]);
printf("%d\n",sum);
}
return 0;
}