贪心-HDU-5742-It's All In The Mind

It’s All In The Mind

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 146 Accepted Submission(s): 73

Problem Description
Professor Zhang has a number sequence a1,a2,…,an. However, the sequence is not complete and some elements are missing. Fortunately, Professor Zhang remembers some properties of the sequence:

1. For every i∈{1,2,…,n}, 0≤ai≤100.
2. The sequence is non-increasing, i.e. a1≥a2≥…≥an.
3. The sum of all elements in the sequence is not zero.

Professor Zhang wants to know the maximum value of (a1+a2)/(a1+…+an) among all the possible sequences.

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains two integers n and m (2≤n≤100,0≤m≤n) – the length of the sequence and the number of known elements.

In the next m lines, each contains two integers xi and yi (1≤xi≤n,0≤yi≤100,xi < xi+1,yi ≥ yi+1), indicating that axi=yi.

Output
For each test case, output the answer as an irreducible fraction “p/q”, where p, q are integers, q>0.

Sample Input
2
2 0
3 1
3 1

Sample Output
1/1
200/201

Author
zimpha

Source
2016 Multi-University Training Contest 2

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题意：
有一个由n个数组成的数字序列，其中只知道m个数（位置和值），并且这是一个单调不增的序列，且数字的最大值为100，现在想要知道(a1+a2)/(a1+…+an)这个值最大能是多少。

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题解：
观察这个值就知道，要让a1,a2尽量大，其余的尽量小就可以啦。

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////  main.cpp//  160721-1009////  Created by 袁子涵 on 16/7/21.//  Copyright © 2016年 袁子涵. All rights reserved.//#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <vector>#include <cmath>#include <queue>#include <algorithm>#include <string>using namespace std;int t,n,m,head,tail,now,fz,fm,nextmax,yue;typedef struct{    int plz,num;}node;node data[105];int cmp(node a,node b){    return a.plz<b.plz;}int gcd(int a,int b){    if (b==0)        return a;    return gcd(b, a%b);}int main(void){    ios::sync_with_stdio(false);    cin >> t;    while (t--) {        cin >> n >> m;        for (int i=0; i<m; i++)            cin >> data[i].plz >> data[i].num;        head=fz=fm=0;        tail=m;        nextmax=100;        now=1;        data[tail].plz=n+1,data[tail].num=0;        sort(data+0, data+m, cmp);        for (int i=0; i<=m; i++) {            while (now<=min(2,data[i].plz-1)) {                fz+=nextmax;                fm+=nextmax;                now++;            }            if (now<=2) {                fz+=data[i].num;                fm+=data[i].num;            }            else                fm+=(data[i].plz+1-now)*data[i].num;            now=data[i].plz+1;            nextmax=data[i].num;        }        yue=gcd(fm,fz);        fm/=yue;        fz/=yue;        cout << fz << '/' << fm << endl;    }    return 0;}