• 2016女生赛 HDU 5710 Digit-Sum(数学,思维题)


    Digit-Sum

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
    Total Submission(s): 782    Accepted Submission(s): 241


    Problem Description
    Let S(N) be digit-sum of N , i.e S(109)=10,S(6)=6 .

    If two positive integers a,b are given, find the least positive integer n satisfying the condition a×S(n)=b×S(2n) .

    If there is no such number then output 0.
     
    Input
    The first line contains the number of test caces T(T10) .
    The next T lines contain two positive integers a,b(0<a,b<101) .
     
    Output
    Output the answer in a new line for each test case.
     
    Sample Input
    3 2 1 4 1 3 4
     
    Sample Output
    1 0 55899
     
    Source
     
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    #include<stdio.h>
    #include<string.h>
    #include<stdio.h>  
    #include<string.h>  
    #include<stdlib.h>  
    #include<queue>  
    #include<stack>  
    #include<math.h>  
    #include<vector>  
    #include<map>  
    #include<set>  
    #include<cmath>  
    #include<complex>  
    #include<string>  
    #include<algorithm>  
    #include<iostream>
    #include<string.h>
    #include<algorithm>
    #include<vector>
    #include<stdio.h>
    #include<cstdio>
    #include<time.h>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<map>
    #define inf 0x3f3f3f3f
    #define ll long long
    using namespace std;
    int d[100005];
    int gcd(int a,int b)
    {
       return b==0?a:gcd(b,a%b);
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            bool ff=0;
            bool f=0;
            int x=2*b-a;
            int y=9*b;
            
            if(x==0)
            {
                cout<<1<<endl;
                continue;
            }
            else if(x<0||5*x>y) 
            {
                cout<<"0"<<endl;
                continue;
            }
            int xx,yy;
            xx=max(x,y);
            yy=min(x,y);
            int pp=gcd(xx,yy);
            x=x/pp;
            y=y/pp;
            y=y-5*x;
            memset(d,0,sizeof(d));
            for(int i=1;i<=x;i++) d[i]=5;
            int i=1;
            while(y>=4)
            {
                y=y-4;
                d[i]+=4;
                i++;
            }
            x=max(x,i-1);
            if(y)
            {
                d[i]+=y;
                if(x==i-1) x++;
            }
            for(int j=x;j>=1;j--)
                cout<<d[j];
            cout<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/caiyishuai/p/9034149.html
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