• cf1060C. Maximum Subrectangle(思维 枚举)


    题意

    题目链接

    Sol

    好好读题 => 送分题

    不好好读题 => 送命题

    开始想了(30)min数据结构发现根本不会做,重新读了一遍题发现是个傻逼题。。。

    (C_{i, j} = a[i] * b[j])

    根据乘法分配律,题目就变成了在数组(a, b)中分别选一段连续的区间,要求权值和相乘(<= X),最大化区间长度乘积

    (n^2)模拟一下即可

    #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;	
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10079774.html
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