51Nod 1305 Pairwise Sum and Divide
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这题看了讨论区的说就与1和2有关
自己真欠缺思考。如果两个数a,b不为1或者2 。a+b/(a*b)=0
如果取一个1和一个不为1的,a+b/(a*b)=1
取两个1,a+b/(a*b)=2
取两个2,a+b/(a*b)=1
然后只要保存1和2的个数就行
#include<iostream> #include<cstdio> #include<algorithm> #include<vector> #include<set> #include<queue> #include<cstring> using namespace std; int main(){int n;cin>>n;int a1=0,a2=0;for(int i=0;i<n;i++){int num;scanf("%d",&num);if(num==1) a1++;if(num==2) a2++;}int sum=0;sum=a1*(n-a1);//取一个1和任意一个不为1 sum+=a1*(a1-1);//取两个1 sum+=a2*(a2-1)/2;//取两个2 cout<<sum;return 0;}
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