简单题,考虑这个串只有0,1两种字符。所以每9个必有一组合法的情况
所以暴力的复杂度是\(O(9n)\)
代码:
#include#include #include #include using namespace std;#define rg registervoid read(int &x){ char ch;bool ok; for(ok=0,ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-')ok=1; for(x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());if(ok)x=-x;}const int maxn=3e5+10;int n,las=1e9,d[maxn];long long ans;char a[maxn];int main(){ scanf("%s",a+1),n=strlen(a+1); for(rg int i=1;i<=n;i++) for(rg int j=1;i+j*2<=n;j++) if(a[i]==a[i+j]&&a[i+j*2]==a[i]){d[i]=i+j*2;break;} for(rg int i=n;i>=1;i--){ if(d[i])las=min(las,d[i]); if(las!=1e9)ans+=n-las+1; } printf("%lld\n",ans);}