【codeforce】A. Group of Students

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本文探讨了如何通过编程算法合理地将学生分为初级和中级编程组，确保每个组的人数在指定范围内。文章详细介绍了输入数据格式、算法实现及复杂度分析，并通过实例展示了解决方案的有效性。

原题：

At the beginning of the school year Berland State University starts two city school programming groups, for beginners and for intermediate coders. The children were tested in order to sort them into groups. According to the results, each student got some score from 1 to m points. We know that c1 schoolchildren got 1 point, c2 children got 2 points, ..., cmchildren got m points. Now you need to set the passing rate k (integer from 1 to m): all schoolchildren who got less than k points go to the beginner group and those who get at strictly least k points go to the intermediate group. We know that if the size of a group is more than y, then the university won't find a room for them. We also know that if a group has less than x schoolchildren, then it is too small and there's no point in having classes with it. So, you need to split all schoolchildren into two groups so that the size of each group was from x to y, inclusive.

Help the university pick the passing rate in a way that meets these requirements.

Input

The first line contains integer m (2 ≤ m ≤ 100). The second line contains m integers c1, c2, ..., cm, separated by single spaces (0 ≤ ci ≤ 100). The third line contains two space-separated integers x and y (1 ≤ x ≤ y ≤ 10000). At least one ci is greater than 0.

Output

If it is impossible to pick a passing rate in a way that makes the size of each resulting groups at least x and at most y, print 0. Otherwise, print an integer from 1 to m — the passing rate you'd like to suggest. If there are multiple possible answers, print any of them.

Sample test(s)

input
5
3 4 3 2 1
6 8

output
3

input
5
0 3 3 4 2
3 10

output
4

input
2
2 5
3 6

output
0

Note

In the first sample the beginner group has 7 students, the intermediate group has 6 of them.

In the second sample another correct answer is 3.

分析：

简单的思维题，先求出总的分数sum和得分不大于ci的人数s[n]，复杂度为O(n)，然后检查s[i]和sum-s[i]是否在[x,y]之内，是的话直接输出i+2，反之输出0，复杂度为O(n).

代码：

#include <iostream>

using namespace std;
int Solution(int num[],int n,int x,int y);
bool CheckSolution(int m,int sum,int x,int y);
int main()
{
    int n;
    cin >> n;

    int num[n];
    for(int i=0 ; i<n ; i++){
        cin >> num[i];
    }

    int x,y;
    cin >> x >> y;

    cout << Solution(num,n,x,y) << endl;
    return 0;
}
int Solution(int num[],int n,int x,int y)
{
    int ret = 0;
    int s[n];
    int sum = num[0];
    s[0] = num[0];

    for(int i=1 ; i<n ; i++){
        sum = sum + num[i];
        s[i] = s[i-1] + num[i];
    }
    for(int i=0 ; i<n ; i++){
        if(true == CheckSolution(s[i],sum,x,y)){
            ret = i+1+1;
            break;
        }
    }
    return ret;
}
bool CheckSolution(int m,int sum,int x,int y)
{
    if(m>=x && m<=y && sum-m>=x && sum-m<=y){
        return true;
    }
    else{
        return false;
    }
}

总结：看懂题目比较简单。