UVA1585 LA3354 Score【水题】

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

问题链接：UVA1585 LA3354 Score
问题简述：（略）
问题分析：
这个题是计算得分，得分规则有点像保龄球以及跳一跳的得分规则，连续正确的次数越多分数越高。
这个题逻辑简单，用3种语言实现。
程序说明：解题代码在LA3354（UVALive3354）中会出现WA，不理解。
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA1585 LA3354 Score */#include <bits/stdc++.h>using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);int t;cin >> t;while(t--) {string s;cin >> s;int ans = 0, cnt = 0;for(int i = 0; s[i]; i++)if(s[i] == 'O') ans += ++cnt;else cnt = 0;cout << ans << endl;}return 0;
}

AC的C语言程序（字符流）如下：

/* UVA1585 LA3354 Score */#include <stdio.h>int main(void)
{int t;scanf("%d", &t);getchar();while(t--) {int ans = 0, cnt = 0;char c;while((c = getchar()) != '\n') {if(c == 'O') ans += ++cnt;else if(c == 'X') cnt = 0;}printf("%d\n", ans);}return 0;
}

AC的C语言程序（格式化输入）如下：

/* UVA1585 LA3354 Score */#include <stdio.h>#define N 80
char s[N + 1];int main(void)
{int t, i;scanf("%d", &t);while(t--) {int ans = 0, cnt = 0;scanf("%s", s);for(i = 0; s[i]; i++)if(s[i] == 'O') ans += ++cnt;else cnt = 0;printf("%d\n", ans);}return 0;
}

AC的Python语言程序如下：

# UVA1585 LA3354 Scoret = int(input())
for k in range(t):a = input()ans = 0cnt = 0for i in a:if i == 'O':cnt += 1ans += cntelse:cnt = 0print(ans)

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