题解 (by;zjvarphi)
压行。
枚举两行,将中间的行压成一行,然后直接前缀和加二分。
注意边界细节问题。
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
",ans);
return 0;
}
}
int main() {return nanfeng::main();}