• hdu 5461(分类讨论)


    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 ij 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 (2n5×106), a (0|a|106) and b (0|b|106). The second line contains n integers t1,t2,,tn where 0|ti|106 for 1in.

    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.
     
    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
     
    Source
     
    昨天输的很惨,中南地区邀请赛只A了两道题。所以感觉要多做少错思路题。
    这个题分4种情况考虑,每种又分为三种情况。举一个例子:
    a>0,b<0时
    那么ti^2尽可能的大,tj尽可能的小。如果i!=j 那么直接取ti^2最大与tj最小。
    如果i==j 那么在ti^2次大与tj最小和ti^2最大与tj次小中取大者。
    还有就是 INF不能这样赋值 ,如1<<40 ,这个地方WA了好多次,,他会变成0
    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include <stdlib.h>
    #include<math.h>
    #include<algorithm>
    using namespace std;
    typedef long long LL;
    const LL INF = 1<<60;
    int main(){
        int tcase;
        scanf("%d",&tcase);
        int t =1;
        while(tcase--){
            int n,a,b;
            scanf("%d%d%d",&n,&a,&b);
            LL ans;
            if(a>=0&&b>=0){
                LL fmax = -INF,smax = -INF;
                for(int i=1;i<=n;i++){
                    LL num;
                    scanf("%lld",&num);
                    if(num>fmax){
                        smax = fmax;
                        fmax = num;
                    }else{
                        smax = max(smax,num);
                    }
                }
                ans = max(a*fmax*fmax+b*smax,a*smax*smax+b*fmax);
            }else if(a>0&&b<0){
                LL MAX=-INF,SMAX=-INF; ///平方的最大和次大
                LL MIN=INF,SMIN=INF; ///最小和次小
                int id1,id2;
                for(int i=1;i<=n;i++){
                    LL num;
                    scanf("%lld",&num);
                    LL temp = num*num;
                    if(num<MIN){
                        SMIN = MIN;
                        MIN = num;
                        id1 = i;
                    }else SMIN = min(SMIN,num);
                    if(temp>MAX){
                        SMAX = MAX;
                        MAX  = temp;
                        id2 = i;
                    }else SMAX = max(SMAX,num);
                }
                if(id1!=id2){
                    ans = a*MAX+b*MIN;
                }else{
                    ans = max(a*SMAX+b*MIN,a*MAX+b*SMIN);
                }
            }else if(a<0&&b>0){
                LL MAX = -INF,MIN = INF; ///注意这里的最小是指平方的最小
                LL smin=INF,smax = -INF; ///此处次小是指平方的次小
                int id1,id2; ///分别记录ti 和 tj 的位置
                for(int i=1;i<=n;i++){
                    LL num;
                    scanf("%lld",&num);
                    if(num>MAX) {
                        smax = MAX;
                        MAX = num;
                        id1 = i;
                    }else smax = max(smax,num);
                    LL temp = num*num;
                    if(temp<MIN){
                        smin = MIN;
                        MIN = temp;
                        id2 = i;
                    }else smin = min(smin,temp);
                }
                if(id1!=id2){
                    ans = a*MIN+b*MAX;
                }
                else{
                    ans = max(a*MIN+b*smax,a*smin+b*MAX);
                }
            }else {
                LL MIN = INF,SMIN = INF; ///这两个值代表绝对值次小和最小的平方
                LL MIN1 = INF,SMIN1 = INF;  ///这两个值代表次小和最小
                int id1,id2;
                for(int i=1;i<=n;i++){
                    LL num;
                    scanf("%lld",&num);
                    LL temp = num*num;
                    if(num<MIN1) {
                        SMIN1 = MIN1;
                        MIN1 = num;
                        id1 = i;
                    }else SMIN1 = min(SMIN1,num);
    
                    if(temp<MIN){
                        SMIN = MIN;
                        MIN = temp;
                        id2 = i;
                    }else SMIN = min(SMIN,temp);
                }
                if(id1!=id2){
                    ans = a*MIN+b*MIN1;
                }else{
                    ans = max(a*MIN+b*SMIN1,a*SMIN+b*MIN1);
                }
            }
            printf("Case #%d: %lld
    ",t++,ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/liyinggang/p/5565513.html
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