题目：
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.
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 .
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
题目大意：有一个数列a，但是很多数字都遗失了，只记得其中一些数字，给出数列的长度，并给出记得的数字信息（数列的第几个是多少），要求求出（a1+a2）/(a1+a2+........an)的最大值（约分后的结果）
解题思路：因为（a1+a2）<(a1+a2+........an)，因此该数值一定是一个小于1的真分数，要使分数尽可能大则应该使分子尽可能大同时分母尽可能小，又根据题目给出的限定条件发现，0<=ai<=100，因此a1和a2应该尽可能从100往下取，而其他值则应该尽可能从0往上取，题目给出的另外一个条件是该数列是一个单调递减的数列,ai>=ai+1，因此可以从数列后部往前遍历，某位置的元素等于后方的不为零的值，这样可以得到满足要求的数组，然后求和，非常重要的一点是要记得进行约分，最后输出约分后的结果
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{int t;int n,m;int x,y;int ansnum=0;int ansden=0;int num[105];scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(num,0,sizeof(num));num[0]=100;num[1]=100;while(m--){scanf("%d%d",&x,&y);num[x-1] = y;}if(num[0]<num[1])num[1] = num[0];for(int i=n-2;i>=2;i--){if((num[i]==0)&&(num[i+1]!=0))num[i] = num[i+1];}ansnum=ansden=0;ansnum += (num[0]+num[1]);for(int i=0;i<n;i++){ansden+=num[i];}/* for(int i=0;i<n;i++){cout<<num[i]<<" ";}cout<<endl;*/for (int i = 2; i <= ansnum; i++){if (ansnum % i == 0 && ansden % i == 0){ansnum /= i;ansden /= i;i--;continue;}}printf("%d/%d\n",ansnum,ansden);}return 0;
}
