題目鏈接:Problem - B - Codeforces

每當出現一次不同時,前面的所有序列就都要再計算一次

ac代碼:

#include <cstdio> #include <iostream> #include <algorithm> #include <queue> #include <cmath> #include <cstring> #include <string> #include <stack> #include <deque> #include <map> #include <set> using namespace std; #define ll long long #define endl "\n" #define rep(i, a, b) for (ll i = (a); i <= (b); i++) #define repr(i, a, b) for (ll i = (a); i < (b); i++) #define rrep(i, a, b) for (ll i = (b); i >= (a); i--) #define rrepr(i, a, b) for (ll i = (b); i > (a); i--) #define min(a,b) (a)<(b)?(a):(b) #define max(a,b) (a)>(b)?(a):(b)ll cnt,n,m,t,ans,ant; const int N=1e5+10; ll arr[N]; string str;inline ll read() {char c = getchar();int x = 0,s = 1;while(c < '0' || c > '9') {if(c == '-') s = -1;c = getchar();}//是符號while(c >= '0' && c <= '9') {x = x*10 + c -'0';c = getchar();}//是數字return x*s; }void solve() {n=read();cin>>str;ans=n;repr(i,1,n)if(str[i-1]!=str[i]) ans+=i;cout<<ans<<endl;return; }int main() {t=read();while(t--)solve();return 0; }

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