題意 "題目鏈接" Sol 好好讀題 = 送分題 不好好讀題 = 送命題 開始想了$30$min數據結構發現根本不會做,重新讀了一遍題發現是個傻逼題。。。 $C_{i, j} = a[i] b[j]$ 根據乘法分配律,題目就變成了在數組$a, b$中分別選一段連續的區間,要求權值和相乘$ defin ...
題意
Sol
好好讀題 => 送分題
不好好讀題 => 送命題
開始想了\(30\)min數據結構發現根本不會做,重新讀了一遍題發現是個傻逼題。。。
\(C_{i, j} = a[i] * b[j]\)
根據乘法分配律,題目就變成了在數組\(a, b\)中分別選一段連續的區間,要求權值和相乘\(<= X\),最大化區間長度乘積
\(n^2\)模擬一下即可
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define rg register
#define pt(x) printf("%d ", x);
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
#define chmin(x, y) (x = x < y ? x : y)
using namespace std;
const int MAXN = 2001, INF = 1e18 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int 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;
}
int N, M, a[MAXN], b[MAXN], f[MAXN], g[MAXN], Lim, ans;
main() {
N = read(); M = read();
for(int i = 1; i <= N; i++) a[i] = read();
for(int i = 1; i <= M; i++) b[i] = read();
memset(f, 0x7f, sizeof(f));
memset(g, 0x7f, sizeof(g));
Lim = read();
for(int i = 1; i <= N; i++) {
int mn = 0;
for(int j = i; j <= N; j++) mn += a[j], chmin(f[j - i + 1], mn);
}
for(int i = 1; i <= M; i++) {
int mn = 0;
for(int j = i; j <= M; j++) mn += b[j], chmin(g[j - i + 1], mn);
}
for(int i = 1; i <= N; i++) {
for(int j = 1; j <= M; j++) {
if(f[i] * g[j] <= Lim)
ans = max(ans, j * i);
}
}
cout << ans;
return 0;
}