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(T≤10)

.
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

"巴卡斯杯"
中国大学生程序设计竞赛 - 女生专场

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liuyiding

#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;
}

原文地址：https://www.cnblogs.com/caiyishuai/p/9034149.html