hdu 5461 Largest Point（杂题）
Largest Point
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1485 Accepted Submission(s): 588
Problem Description
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
at2i+btj, becomes the largest point.
Input
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 nintegers 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.
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 nintegers 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.
Output
The output contains exactly
T lines.
For each test case, you should output the maximum value of at2i+btj.
For each test case, you should output the maximum value of at2i+btj.
Sample Input
2 3 2 1 1 2 3 5 -1 0 -3 -3 0 3 3
Sample Output
Case #1: 20 Case #2: 0
思路：分四种情况讨论即可。
代码：
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
#define INF 999999999
int main(){
int tcase;
scanf("%d",&tcase);
int t =1;
while(tcase--){
LL IMAX=-INF,ISMAX=-INF,IMIN=INF,ISMIN=INF; ///ti平方的最大次大最小次小
LL JMAX=-INF,JSMAX=-INF,JMIN=INF,JSMIN=INF; ///tj的最大次大最小次小
int n;
LL a,b;
int id1,id2,id3,id4; ///ti 最大最小 tj最大最小
scanf("%d%lld%lld",&n,&a,&b);
for(int i=1;i<=n;i++){
LL NUM,TEMP;
scanf("%lld",&NUM);
TEMP = NUM*NUM;
if(NUMJMAX){
JSMAX = JMAX;
JMAX = NUM;
id1 = i;
}else JSMAX = max(JSMAX,NUM);
if(TEMPIMAX){
ISMAX = IMAX;
IMAX = TEMP;
id3 = i;
}else ISMAX = max(ISMAX,TEMP);
}
LL ans = 0;
if(a>=0&&b>=0){
if(id3!=id1) ans = a*IMAX+b*JMAX;
else ans = max(a*ISMAX+b*JMAX,a*IMAX+b*JSMAX);
}else if(a>0&&b<0){
if(id3!=id2) ans = a*IMAX+b*JMIN;
else ans = max(a*ISMAX+b*JMIN,a*IMAX+b*JSMIN);
}else if(a<0&&b>0){
if(id4!=id1) ans = a*IMIN+b*JMAX;
else ans = max(a*ISMIN+b*JMAX,a*IMIN+b*JSMAX);
}else {
if(id4!=id2) ans = a*IMIN+b*JMIN;
else ans = max(a*IMIN+b*JSMIN,a*ISMIN+b*JMIN);
}
printf("Case #%d: %lld\n",t++,ans);
}
return 0;
}