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.
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) .
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; }