[codeforces] Gym - 100814E Everything Has Changed (DP)

[codeforces] Gym - 100814E Everything Has Changed (DP)

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題目鏈接：
Everything Has Changed

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題目大意：
給你一個n∗m 的矩陣， 讓你從左上角走到右下角 只能往下或者往右選擇一些數字相乘， 使得最後的乘積轉換爲6進制之後0的個數儘可能多。

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數據範圍：
2≤n,m≤100

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解題思路：
要想一個數轉換爲六進制0儘可能多， 那麼我們需要儘可能讓一個數裏2， 3的個數儘可能多。 所以我們可以先處理出每個數包含多少個2， 多少個3，之後再去DP處理， 我用dp[i][j][k] 表示i 行j 列選k 個3最多選幾個2

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;
const int MaxN = 100;

struct NODE{
    int c2, c3;
}a[MaxN + 5][MaxN + 5];

int dp[MaxN +5][MaxN + 5][MaxN * 20 + 5];
int n, m, T;

void solve(){
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            for(int k = a[i][j].c3; k <= 2000; k++){
                if(dp[i - 1][j][k - a[i][j].c3] != -1)
                    dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j][k - a[i][j].c3] + a[i][j].c2);
                if(dp[i][j - 1][k - a[i][j].c3] != -1)
                    dp[i][j][k] = max(dp[i][j][k], dp[i][j - 1][k - a[i][j].c3] + a[i][j].c2);
            }
}

int main()
{
    scanf("%d", &T);
    while(T--){
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                int x;
                scanf("%d", &x);
                a[i][j].c2 = a[i][j].c3 = 0;
                while(x % 2 == 0) a[i][j].c2++, x /= 2;
                while(x % 3 == 0) a[i][j].c3++, x /= 3;
            }
        }
        memset(dp, -1, sizeof(dp));
        for(int i = 1; i <= n; i++) dp[i][0][0] = 0;
        for(int j = 1; j <= m; j++) dp[0][j][0] = 0;
        solve();
        int ans = 0;
        for(int k = 0; k <= 2000; k++) ans = max(ans, min(k, dp[n][m][k]));
        printf("%d\n", ans);
    }
    return 0;
}

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標籤（空格分隔）： DP codeforces