• 1384. 饲料调配


    枚举。每种饲料的份数取值为(0 sim 100),时间复杂度为(O(101^3))

    int g[3];
    int a[3][3];
    
    int main()
    {
        for(int i = 0; i < 3; i++) cin >> g[i];
        if(g[0] + g[1] + g[2] == 0) puts("0 0 0 0");
        else
        {
            for(int i = 0; i < 3; i++)
                for(int j = 0; j < 3; j++)
                    cin >> a[i][j];
            
            int sum = 301;
            int ans[4] = {0};
            for(int i = 0; i <= 100; i++)
                for(int j = 0; j <= 100; j++)
                    for(int k = 0; k <= 100; k++)
                    {
                        if(i + j + k == 0) continue;
                        int s[3] = {0};
                        for(int l = 0; l < 3; l++)
                            s[l] = i * a[0][l] + j * a[1][l] + k * a[2][l];
                        
                        int t = (s[0] + s[1] + s[2]) / (g[0] + g[1] + g[2]);
                        if(s[0] == t * g[0] && s[1] == t * g[1] && s[2] == t * g[2] && i + j + k < sum)
                        {
                            sum = i + j + k;
                            ans[0] = i, ans[1] = j, ans[2] = k, ans[3] = t;
                        }
                    }
            
            if(sum > 300) puts("NONE");
            else 
            {
                for(int i = 0; i < 4; i++)
                    cout << ans[i] << ' ';
                cout << endl;
            }
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/fxh0707/p/14897748.html
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