Codeforces Round 513 (Div.1+Div.2)
10月4号的比赛,因为各种原因(主要是懒),今天才写总结……
Div1+Div2,只做出两个题+迟到\(20min\),日常掉\(rating\)……
\(\rm{A.Phone\;Numbers}\)
很水的一道题目,直接输出cout<<min(n/11,tot);
(\(tot\)为数字\(8\)的数量)
\(\mathcal{Maximum\;Sum\;of\;Digits}\)
一点小贪心,我们让\(9\)最多就行了,证明吗……感性的理解一下吧
\(\mathfrak{Maximum\;Subrectangle}\)
因为矩阵的元素\(C_{i,j}=a_i\times b_i\),所以
\(\sum_{i=x_1}^{x2}{\sum_{j=y_1}^{y2}{C_{i,j}}}=(sumx[x_2]-sumx[x_1-1])\times (sumy[x_2]-sumy[x_1-1])\)
其中sumx
和sumy
是\(a\)数列和\(b\)数列的前缀和。这样我们就可以\(O(n^2)\)处理出子矩阵长一定时候的最小值和宽一定时的最小值,然后\(O(1)\)判断。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long sumx[2010],sumy[2010],minx[2010],miny[2010];
long long read(){
int k=0,f=1; char c=getchar();
for(;c<'0'||c>'9';c=getchar())
if(c=='-') f=-1;
for(;c>='0'&&c<='9';c=getchar())
k=k*10+c-48;
return k*f;
}
int main(){
int n=read(),m=read();
memset(minx,127,sizeof(minx)), memset(miny,127,sizeof(miny));
for(int i=1;i<=n;i++) sumx[i]=sumx[i-1]+read();
for(int i=1;i<=m;i++) sumy[i]=sumy[i-1]+read();
//====预处理长宽一定时的最小值
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
minx[i]=min(minx[i],sumx[j]-sumx[j-i]);
}
}
//====判断并记录ans
for(int i=1;i<=m;i++){
for(int j=i;j<=m;j++){
miny[i]=min(miny[i],sumy[j]-sumy[j-i]);
}
}
int x=read(),ans=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++)
if(minx[i]*miny[j]<=x) ans=max(ans,i*j);
}
cout<<ans;
return 0;
}
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