2016女生赛HDU5710DigitSum（数学，思维题）

Z时代
2024-01-10
分类：综合

coding

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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#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
usingnamespace 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;
        }
elseif(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;
    }
return0;}