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最新推荐文章于 2019-02-02 00:21:00 发布

原创最新推荐文章于 2019-02-02 00:21:00 发布 · 654 阅读

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本文介绍了一种针对客观题答案进行计分的算法，并提供了一个具体的C++实现方案。该算法通过跟踪连续正确的答案来计算分数，适用于竞赛或考试等场景。文中详细解释了动态规划方法的应用，包括状态转移方程。

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There is an objective test result such as OOXXOXXOOO". An `O' means a correct answer of a problem and an `X' means a wrong answer. The score of each problem of this test is calculated by itself and its just previous consecutive `O's only when the answer is correct. For example, the score of the 10th problem is 3 that is obtained by itself and its two previous consecutive `O's. Therefore, the score ofOOXXOXXOOO” is 10 which is calculated by `1+2+0+0+1+0+0+1+2+3". You are to write a program calculating the scores of test results. Input Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing a string composed by O’ and X' and the length of the string is more than 0 and less than 80. There is no spaces between O’ and ` X’.
Output
Your program is to write to standard output. Print exactly one line for each test case. The line is to contain the score of the test case.
The following shows sample input and output for five test cases.
Sample Input
5
OOXXOXXOOO
OOXXOOXXOO
OXOXOXOXOXOXOX
OOOOOOOOOO
OOOOXOOOOXOOOOX
Sample Output
10
9
7
55
30
0是对，x是错。0是根据连续得分的，000，第一个是1分，第二个是2分，第三个是3分，但是一旦错了，从1再继续计。
dp[i]=dp[i-1]+1;
dp[i]=0;
就是这两个状态。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<queue>
#include<algorithm>
using namespace std;
char a[100];
int dp[100];
int main()
{
    int i, j, m, n, ans, t, len;
    cin >> t;
    getchar();
    while (t--)
    {
        gets(a);
        memset(dp, 0, sizeof(dp));
        len = strlen(a);
        if (a[0] == 'O')dp[0] = 1;
        else dp[0] = 0;
        for (i = 1; i < len; i++)
        {
            if (a[i] == 'O')
                dp[i] = dp[i - 1] + 1;
            else dp[i] = 0;
        }
        ans = 0;
        for (i = 0; i < len; i++)
            ans = ans + dp[i];
        cout << ans << endl;
    }

    return 0;
}