Description
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.
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
本題的思路是，定義一個二層循環，第一層循環從左往右找，一旦找到字符”O“，就進入第二層循環，第二層循環依然從左往右搜索，直到遇到“X”或者到了最右邊的邊界爲止。然後把j賦值給i，返回第一層循環。</pre><pre code_snippet_id="1654856" snippet_file_name="blog_20160420_1_3933707" name="code" class="cpp">#include <string>
#include <iostream>
using namespace std;
int main(int argc, const char * argv[]) {
int T;
cin >> T;//變量T表示T組數據
string s;//保存字符串
while(T--) {
cin >> s;
int sum = 0; //保存結果
for (int i = 0; i < s.length(); i++) {
if (s[i] != 'O') continue; //如果字符不爲'O',直接跳到下次循環。
int tempSum = 0;
for (int j = i; j < s.length(); j++) {
if (s[j] == 'O') {
tempSum++;
sum = sum + tempSum;
i = j;
}else {
i = j;
break;
}
}
}
cout << sum << endl;
}
return 0;
}
