NOIP 模擬 $38\; \rm a$
阿新 • • 發佈:2021-08-13
題解
題解 \(by\;zj\varphi\)
壓行。
列舉兩行,將中間的行壓成一行,然後直接字首和加二分。
注意邊界細節問題。
Code
#include<bits/stdc++.h> #define Re register #define ri Re signed #define p(i) ++i namespace IO{ char buf[1<<21],*p1=buf,*p2=buf; #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++ struct nanfeng_stream{ template<typename T>inline nanfeng_stream &operator>>(T &x) { Re bool f=0;x=0;Re char ch=getchar(); while(!isdigit(ch)) f|=ch=='-',ch=getchar(); while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar(); return x=f?-x:x,*this; } }cin; } using IO::cin; namespace nanfeng{ #define FI FILE *IN #define FO FILE *OUT template<typename T>inline T cmax(T x, T y) {return x>y?x:y;} template<typename T>inline T cmin(T x, T y) {return x>y?y:x;} typedef long long ll; static const int N=5e4+7; int mat[33][N],sum[33][N],pre[N],n,m,l,r; char s[N]; ll ans; inline int main() { //FI=freopen("nanfeng.in","r",stdin); //FO=freopen("nanfeng.out","w",stdout); cin >> n >> m; for (ri i(1);i<=n;p(i)) { scanf("%s",s+1); for (ri j(1);j<=m;p(j)) mat[i][j]=s[j]=='1',sum[i][j]=sum[i-1][j]+mat[i][j]; } cin >> l >> r; for (ri i(1);i<=n;p(i)) for (ri j(1);j<=i;p(j)) for (ri k(1);k<=m;p(k)) { pre[k]=pre[k-1]+sum[i][k]-sum[j-1][k]; if (pre[k]<l) continue; int le=pre[k]-l; int nm=std::upper_bound(pre,pre+k,le)-pre; --nm; ans+=nm+1; if (pre[k]<=r) continue; int re=pre[k]-r; nm=std::lower_bound(pre,pre+k,re)-pre; ans-=nm; } printf("%lld\n",ans); return 0; } } int main() {return nanfeng::main();}