本文主要是介绍hdu 5461 Largest Point，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Given the sequence A A with n n integers t 1 ,t 2 ,⋯,t n t1,t2,⋯,tn. Given the integral coefficients a a and b b. The fact that select two elements t i ti and t j tj of A A and i≠j i≠j to maximize the value of at 2 i +bt j ati2+btj, becomes the largest point.
For each test case, the first line contains three integers corresponding to n (2≤n≤5×10 6 ), a (0≤|a|≤10 6 ) n (2≤n≤5×106), a (0≤|a|≤106) and b (0≤|b|≤10 6 ) b (0≤|b|≤106). The second line contains n n integers t 1 ,t 2 ,⋯,t n t1,t2,⋯,tn where 0≤|t i |≤10 6 0≤|ti|≤106 for 1≤i≤n 1≤i≤n.
The sum of n n for all cases would not be larger than 5×10 6 5×106.
For each test case, you should output the maximum value of at 2 i +bt j ati2+btj.
23 2 1 1 2 35 -1 0 -3 -3 0 3 3
Case #1: 20
Case #2: 0
按照ab的符号分好四种情况就可以了
ac代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long a1[5000005];
int main()
{long long n,a,b,t;cin>>t;int p=0;while(t--){p++;cin>>n>>a>>b;cin>>a1[0];long long max1=abs(a1[0]),min1=abs(a1[0]),max2=a1[0],min2=a1[0];if(a<=0&&b<=0){int k=0;for(int i=1;i<n;i++){scanf("%lld",&a1[i]);if(abs(a1[i])<min1){min1=abs(a1[i]);k=i;}}int flag=0;for(int i=1;i<n;i++){if(a1[i]<min2&&i!=k){min2=a1[i];}}printf("Case #%d: %lld\n",p,a*min1*min1+b*min2);}if(a<=0&&b>0){int k=0;for(int i=1;i<n;i++){scanf("%lld",&a1[i]);if(abs(a1[i])<min1){min1=abs(a1[i]);k=i;}}for(int i=1;i<n;i++)if(a1[i]>max2&&i!=k)max2=a1[i];printf("Case #%d: %lld\n",p,a*min1*min1+b*max2);}if(a>0&&b<=0){int k=0;for(int i=1;i<n;i++){scanf("%lld",&a1[i]);if(abs(a1[i])>max1){max1=abs(a1[i]);k=i;}}for(int i=1;i<n;i++){if(a1[i]<min2&&i!=k)min2=a1[i];}printf("Case #%d: %lld\n",p,a*max1*max1+b*min2);}if(a>0&&b>0){int k=0;for(int i=1;i<n;i++){scanf("%lld",&a1[i]);if(abs(a1[i])>max1){max1=abs(a1[i]);k=i;}}for(int i=1;i<n;i++){if(a1[i]>max2&&i!=k)max2=a1[i];}printf("Case #%d: %lld\n",p,a*max1*max1+b*max2);}}return 0;
}
这篇关于hdu 5461 Largest Point的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
