HDU5742 It's All In The Mind(貪心)

HDU5742 It’s All In The Mind(貪心)

鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=5742

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題目

Time Limit:1000MS Memory Limit:65536KB
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∑ni=1ai 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

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題意

給你一個數列的長度和其中某幾個元素，求滿足非升序列的條件下a1+a2∑ni=1ai 的最大值

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分析

最大值即分母儘可能小，a1 和a2 儘可能大。記錄元素位置然後從後往前處理即可保證非升且元素和最小。再對a1 和a2 特別處理一下即可。

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源碼

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<sstream>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
#define mem0(x) memset(x,0,sizeof x)
#define mem1(x) memset(x,-1,sizeof x)
#define dbug cout<<"here"<<endl;
//#define LOCAL

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e6+10;
const int MOD = 1000000007;

int a[110];

int gcd(int a, int b){
    if(b == 0)
        return a;
    return gcd(b, a%b);
}

int main(){
    #ifdef LOCAL
        freopen("C:\\Users\\asus-z\\Desktop\\input.txt","r",stdin);
        freopen("C:\\Users\\asus-z\\Desktop\\output.txt","w",stdout);
    #endif
    int t;
    scanf("%d", &t);
    int n,m;
    while(t--){
        scanf("%d%d", &n, &m);
        mem0(a);
        bool vis[3];
        mem0(vis);
        int tmpP,tmpVal;
        for(int i = 1; i <=m; ++i){
            scanf("%d%d", &tmpP, &tmpVal);
            a[tmpP] = tmpVal;
            if(tmpP==1 || tmpP==2)
                vis[tmpP] = 1;
        }
        ll sum = 0;
        if(!vis[1]){
            a[1] = 100;
            if(!vis[2])
                a[2] = a[1];
        }
        else{
            if(!vis[2])
                a[2] = a[1];
        }
        int lastVal = 0;
        for(int i = n; i >= 1; --i){
            if(!a[i])
                sum += lastVal;
            else{
                lastVal = a[i];
                sum += lastVal;
            }
        }
        int g = gcd(a[1]+a[2], sum);
        printf("%d/%d\n", (a[1]+a[2])/g, sum/g);
    }
    return 0;
}