HDU5742 It's All In The Mind(贪心)

HDU5742 It’s All In The Mind(贪心)

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

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题目

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:

For every i∈{1,2,...,n},0≤ai≤100 . The sequence is non-increasing, i.e. a1≥a2≥...≥an . 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

题意

给你一个数列的长度和其中某几个元素，求满足非升序列的条件下a1+a2∑ni=1ai的最大值

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分析

最大值即分母尽可能小， a1 和 a2 尽可能大。记录元素位置然后从后往前处理即可保证非升且元素和最小。再对 a1 和 a2 特别处理一下即可。

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源码

#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; }