• 杭电2803--The MAX(分析题)


    The MAX

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2033    Accepted Submission(s): 878


    Problem Description
    Giving N integers, V1, V2,,,,Vn, you should find the biggest value of F.
     
    Input
    Each test case contains a single integer N (1<=N<=100). The next line contains N integers, meaning the value of V1, V2....Vn.(1<= Vi <=10^8).The input is terminated by a set starting with N = 0. This set should not be processed.
     
    Output
    For each test case, output the biggest value of F you can find on a line.
     
    Sample Input
    2 1 2 0
     
    Sample Output
    4017
     
    Source
     
    Recommend
    lcy   |   We have carefully selected several similar problems for you:  2807 2802 2804 2800 2816 
    图上给了很多信息, 然而我还是粗心没有看出Xi 可以取相同的数, WA两次。
    #include <cstdio>
    #include <iostream>
    #include <algorithm>
    using namespace std; 
    bool cMp(int a, int b){
        return a > b;
    }
    int main(){
        __int64 num[101];
        int T;
        while(~scanf("%d", &T), T){
            for(int i = 0; i < T; i++)
                scanf("%I64d", &num[i]);
            sort(num, num+T, cMp);
        //    int P = (T-1)%2==0? (T-1)/2*T:T/2*(T-1);
        //    int Q = 2009 - P;
        //    __int64 sum = num[0]*Q;
            __int64 sum = num[0]*(2010-T);
        //    printf("%I64d
    ", sum);
            for(int i = 1; i < T; i++)
                sum += num[i];
            printf("%I64d
    ", sum);
        }
        return 0;
    }  
     
     
  • 相关阅读:
    HTTP协议抓包分析
    cmd 中使用 tracert
    Ubuntu 搭建zabbix
    kerberos+ldap
    运行程序显示丢失“MSVCR100D.dll”
    熊猫烧香病毒样本分析
    Masm32sdk安装指南
    16位汇编实现三大基本排序
    逆向工程初步160个crackme-------3
    一个入门级CTF的Reverse
  • 原文地址:https://www.cnblogs.com/soTired/p/4927463.html
Copyright © 2020-2023  润新知