UVA 1585 Score

最新推荐文章于 2021-02-05 21:30:13 发布

转载最新推荐文章于 2021-02-05 21:30:13 发布 · 100 阅读

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原文链接：http://www.cnblogs.com/Chizhao/p/10439796.html

本文介绍了一个简单的算法，用于计算一种特殊格式的客观测试答案的得分。在这个测试中，正确答案'O'可以累积分数，而错误答案'X'则会重置累积状态。文章提供了完整的C语言实现代码，并通过示例输入输出展示了如何使用该算法。

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 of “OOXXOXXOOO” 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.
Sample Input
5
OOXXOXXOOO
OOXXOOXXOO
OXOXOXOXOXOXOX
OOOOOOOOOO
OOOOXOOOOXOOOOX
Sample Output
10
9
7
55
30
解题思路：在一个只有O和X的序列中，X代表0，连续的O代表首项为1公差为1的等差数列，如果被X切断，那么从首项就变成1，重新开始递增。求被X分割的所有等差数列an的和。

#include <stdio.h>
#include<string.h>
#define MAX 80

int main()
{
    /* code */
    int T,num,len; 
    scanf("%d",&T);
    char str[MAX];
    while(T--){
    num=1;
    scanf("%s",str);
    len=strlen(str);
    int sum=0;//定义初始总和为0
    for (int i = 0; i < len; i++)
    {
             /*等差数列求和*/
        if (str[i]=='O')
        {                

            sum+=num;
            num++;
        }
        else
            num=1;


    }
        printf("%d\n",sum);


    }
    return 0;
}

转载于:https://www.cnblogs.com/Chizhao/p/10439796.html