A Count Task(求選取子串的方案數(這子串是連續的),子串中只有一種元素)
A Count Task
時間限制: 1 Sec 記憶體限制: 128 MB
題目描述
Count is one of WNJXYK’s favorite tasks. Recently, he had a very long string and he wondered that how many substrings which contains exactly one kind of lowercase in this long string. But this string is so long that he had kept counting for several days. His friend Kayaking wants to help him, so he turns to you for help.
輸入
The input starts with one line contains exactly one positive integer T which is the number of test cases.
Each test case contains one line with a string which you need to do a counting task on.
輸出
For each test case, output one line containing “y” where y is the number of target substrings.
樣例輸入
複製樣例資料
3 qwertyuiop qqwweerrttyyuuiioopp aaaaaaaaaa
樣例輸出
10 30 55
提示
1≤T≤20,1≤len(String)≤10^5,1≤∑len(string)≤10^6
Strings only contain lowercase English letters.
比如aaaa4個a,那麼選一個有4種,選兩個有3種,選。。。(連續的!!)和為4+3+2+1,就這樣遍歷一遍就ok了。
/**/ #include <cstdio> #include <cstring> #include <cmath> #include <cctype> #include <iostream> #include <algorithm> #include <map> #include <set> #include <vector> #include <string> #include <stack> #include <queue> typedef long long LL; using namespace std; int n; char s[100005]; int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); scanf("%d", &n); while(n--){ scanf("%s", s); int len = strlen(s); LL ans = 0; for (int i = 0; i < len; i++){ int num = 1; char ch = s[i]; for (i++; i < len; i++){ if(ch == s[i]) num++; else break; } ans += (LL)(1 + num) * num / 2;//記得long long,不然錯的。 i--; } printf("%lld\n", ans); } return 0; } /**/