• hdu 3573(数学+贪心)


    Buy Sticks

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 868    Accepted Submission(s): 392


    Problem Description
    Imyourgod need 3 kinds of sticks which have different sizes: 20cm, 28cm and 32cm. However the shop only sell 75-centimeter-long sticks. So he have to cut off the long stick. How many sticks he must buy at least.
     
    Input
    The first line of input contains a number t, which means there are t cases of the test data.
    There will be several test cases in the problem, each in one line. Each test cases are described by 3 non-negtive-integers separated by one space representing the number of sticks of 20cm, 28cm and 32cm. All numbers are less than 10^6.
     
    Output
    The output contains one line for each line in the input case. This line contains the minimal number of 75-centimeter-long sticks he must buy. Format are shown as Sample Output.
     
    Sample Input
    2 3 1 1 4 2 2
     
    Sample Output
    Case 1: 2 Case 2: 3
     
    Author
    imyourgod (Special Thanks : crackerwang & Louty)
     
    Source
     
    题意:需要 x 根20cm,y根 28cm,z根 32cm,问最少需要多少根 75 cm的木棍才可以都分出来?
    题解:贪心的去选,先选3根的,选不了了才能选2根的,最后如果还有剩才能选一根的.
    这些情况分别是
    20 20 28
    20 20 32
    20 20 20
    32 28
    28 28
    32 32
    20
    28
    32
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #define LL long long
    using namespace std;
    int main()
    {
        int tcase,t = 1;
        scanf("%d",&tcase);
        while(tcase--)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            int ans = 0,a,b,c;
            if(x>=2&&y>=1){ /// 20 20 28
                a = x/2,b=y;
                c = min(a,b);
                x=x-2*c,y=y-c;
                ans+=c;
            }
            if(x>=2&&z>=1){ ///20 20 32
                a = x/2,b = z;
                c = min(a,b);
                x=x-2*c,z=z-c;
                ans+=c;
            }
            if(x>=3){ ///20 20 20
                a = x/3;
                x = x%3;
                ans+=a;
            }
            if(y>=1&&z>=1){ ///28 32
                c = min(y,z);
                ans+=c;
                y-=c;
                z-=c;
            }
            if(y>=2){ ///28 28
                c = y/2;
                y = y%2;
                ans+=c;
            }
            if(z>=2){///32 32
                c = z/2;
                z = z%2;
                ans+=c;
            }
            if(x||y||z) ans++;
            printf("Case %d: %d
    ",t++,ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/liyinggang/p/5917865.html
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