LARGEST POINT
Given the sequence A with n integers t1 , t2 , …… , tn . Given the integral coefficients a and b. The fact that select two elements ti and tj of A and i ≠ j to maximize the value of becomes the largest point.
輸入
An positive integer T, indicating there are T test cases. For each test case, the first line contains three integers corresponding to n (2 ≤ n ≤ 5×106 ), a (0 ≤ |a| ≤ 106 ) and b (0 ≤ |b| ≤ 106 ). The second line contains n integers t1 , t2 , …… , tn where 0 ≤ |ti | ≤ 106 for 1 ≤ i ≤ n.
The sum of n for all cases would not be larger than 5 × 106 .
輸出
The output contains exactly T lines. For each test case, you should output the maximum value of
樣例輸入
2 3 2 1 1 2 3 5 -1 0 -3 -3 0 3 3
樣例輸出
Case #1: 20 Case #2: 0
題意:給定數 n , a, b ,還有接下來的n個ti 的值,求解的最大的值;
分析: 因為 i 和 j 不可以取相等的值,所以將 和 的值分別貪心來求最大的值 ,如果 和 最大值中 i 和 j
的下標相等,則就判斷第二大
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define ll long long int #define maxn 1000010 typedef struct nodee{ ll x,y; }node; node maze[maxn],num[maxn]; bool check(node u,node v) { return u.y<v.y; } int main() { int f,countt=1; ll n,a,b,z; scanf("%d",&f); while(f--){ scanf("%lld %lld %lld",&n,&a,&b); for(int i=0;i<n;i++){ scanf("%lld",&z); maze[i].x=i+1; maze[i].y=a*z*z; num[i].x=i+1; num[i].y=b*z; } printf("Case #%d: ",countt); countt++; sort(maze,maze+n,check); sort(num,num+n,check); if(maze[n-1].x!=num[n-1].x){ printf("%lld\n",maze[n-1].y+num[n-1].y); continue; } ll sum,ans; sum=maze[n-1].y+num[n-2].y; ans=maze[n-2].y+num[n-1].y; printf("%lld\n",max(sum,ans)); } return 0; }