zoukankan      html  css  js  c++  java
  • Holding Bin-Laden Captive!

    We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China!
    “Oh, God! How terrible! ”
     


    Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
    Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
    “Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
    You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!

    InputInput contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.
    OutputOutput the minimum positive value that one cannot pay with given coins, one line for one case.
    Sample Input

    1 1 3
    0 0 0

    Sample Output

    4
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<stack>
    #include<queue>
    #include<iostream>
    #include<map>
    #include<vector>
    #define Inf 0x3f3f3f3f
    #define PI acos(-1.0)
    using namespace std;
    typedef long long ll;
    const int MXAN=300000+10;
    int dp[1234];
    int str[1234];
    int main()
    {
        ll m,n,p,i,j,pos;
        ll a[MXAN];
        ll b[MXAN];
        ll value[4]={0,1,2,5};
        ll num[4];
        while(cin>>num[1]>>num[2]>>num[3])
        {
            if(num[1]+num[2]+num[3]==0)
                break;
            ll m=num[1]*value[1]+num[2]*value[2]+num[3]*value[3];
            for(ll i=0;i<=m;i++)
            {
                a[i]=1;
                b[i]=0;
            }
            a[m+1]=0;
            ll len=value[1]*num[1];
            for(ll i=2;i<=3;i++)
            {
                len+=value[i]*num[i];
                for(ll j=0;j<=len-value[i]*num[i];j++)
                {
                    for(ll k=0;k+j<=len;k+=value[i])
                    {
                        b[k+j]+=a[j];
                    }
                }
                for(ll j=0;j<=len;j++)
                {
                    a[j]=b[j];
                    b[j]=0;
                }
            }
            for(ll i=0;i<=10200;i++)
            {
                if(a[i]==0)
                {
                    cout<<i<<endl;
                    break;
                }
            }
    
        }
        return 0;
    }


  • 相关阅读:
    Asp.net web api部署在某些服务器上老是404
    log4net记录日志到数据库自定义字段
    PIE-Basic 投影变换
    PIE-Basic 波段合成
    PIE-Basic 空间量测
    PIE-Basic 存储格式转换
    PIE-Basic 位深转换
    PIE-Basic 数据拉伸与显示
    PIE-Basic数据信息查看
    PIE-Basic 影像格式转换
  • 原文地址:https://www.cnblogs.com/moomcake/p/9385895.html
Copyright © 2011-2022 走看看