标签：%s scanf strlen 端点 string 分享 namespace base vba
挺神仙的。首先$60$分暴力是比较好打的。
就是枚举左端点，看右端点能否是$0$
但是这样肯定是过不了的，假如我们只枚举一次，把得到的栈记录下来
那么若区间$(l, r)$是可行的，那么$s_{l - } = s_r$，证明自己yy一下吧。。
然后就是字符串hash乱搞了。。
#include<cstdio> #include<algorithm> #include<queue> #include<cstring> #include<map> #define LL long long #define ull unsigned long long using namespace std; const int MAXN = 1e6 + 10, mod = 1e9 + 7; inline LL read() { char c = getchar(); LL x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } char a[MAXN], s[MAXN]; int top = 0; ull base = 12, base2 = 233, po1[MAXN], po2[MAXN]; map<ull, LL> mp; int main() { scanf("%s", a + 1); LL N = strlen(a + 1), ans = 0; ull sum = 0, sum2 = 0; mp[0] = 1; po1[0] = 1; po2[0] = 1; for(int i = 1; i <= N; i++) po1[i] = po1[i - 1] * base, po2[i] = po2[i - 1] * base2; for(int i = 1; i <= N; i++) { if(top && a[i] == s[top]) { top--; sum -= po1[top] * (a[i] - ‘a‘ + 1); sum2 -= po2[top] * (a[i] - ‘a‘ + 1); } else { sum += po1[top] * (a[i] - ‘a‘ + 1); sum2 += po2[top] * (a[i] - ‘a‘ + 1) ; s[++top] = a[i]; } ans += mp[(sum << 5) + sum2]; mp[(sum << 5) + sum2]++; // printf("%d\n", ans); } printf("%lld", ans); return 0; } /* abaababababbbbbaavbaaaabbbaaaabbabbbaabbabbb */
标签：%s scanf strlen 端点 string 分享 namespace base vba
原文地址：https://www.cnblogs.com/zwfymqz/p/9633169.html
