HDU 5742 It's All In The Mind

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∑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

22 03 13 1

 

Sample Output

1/1200/201
尽量让前两个大，后面的小即可。
#include<set>#include<map>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const int N = 1e2 + 10;const int mod = 1e9 + 7;const int INF = 0x7FFFFFFF;int T, n, m, f[N], x, y, a, b;int gcd(int x, int y) { return x%y ? gcd(y, x%y) : y; }int main(){    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n, &m);        memset(f, -1, sizeof(f));        a = b = 0;        while (m--)        {            scanf("%d%d", &x, &y);            f[x] = y;        }        if (f[1] == -1) f[1] = 100;        if (f[2] == -1) f[2] = f[1];        a = f[1] + f[2];        per(i, n, 1)        {            if (f[i] == -1) f[i] = 0;            f[i] = max(f[i], f[i + 1]);            b += f[i];        }        printf("%d/%d\n", a / gcd(a, b), b / gcd(a, b));    }    return 0;}