UVa 1585 / UVALive 3354 Score（字符串的动态规划）

Score

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

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

大致题意：给出一个由O和X组成的串（长度为1~80），统计得分。每个O有一个分，为他前面连续的O个数+1，求所有O的分数和，X没有分。

思路：：动态规划的简单题，类似于LIS，记录O出现几次。

#include<iostream>
#include<algorithm>
#include<string.h>
#include<cstdio>
using namespace std;
char str[100];
int dp[110];
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",str);
        dp[0]=(str[0]=='O');
        for(int i=1;str[i];++i)
        {
            dp[i]=dp[i-1]+1;
            if(str[i]=='X')
                dp[i]=0;
        }
        int sum=0;
        for(int i=0;str[i];++i)
            sum+=dp[i];
        printf("%d\n",sum);
    }
    return 0;
}