Codeforces Beta Round #64 C. Lucky Tickets-优快云博客

本文探讨了在Walrusland公共运输系统中如何通过优化系列数和每系列的票数来确保至少发行一定数量的幸运船票，同时最小化总的发行票数。详细介绍了计算方法及实现步骤。

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In Walrusland public transport tickets are characterized by two integers: by the number of the series and by the number of the ticket in the series. Let the series number be represented by a and the ticket number — by b, then a ticket is described by the ordered pair of numbers (a, b).

The walruses believe that a ticket is lucky if a * b = rev(a) * rev(b). The function rev(x) reverses a number written in the decimal system, at that the leading zeroes disappear. For example, rev(12343) = 34321, rev(1200) = 21.

The Public Transport Management Committee wants to release x series, each containing y tickets, so that at least w lucky tickets were released and the total number of released tickets (x * y) were minimum. The series are numbered from 1 to x inclusive. The tickets in each series are numbered from 1 to y inclusive. The Transport Committee cannot release more than maxx series and more than maxytickets in one series.

Input

The first line contains three integers maxx, maxy, w (1 ≤ maxx, maxy ≤ 105, 1 ≤ w ≤ 107).

Output

Print on a single line two space-separated numbers, the x and the y. If there are several possible variants, print any of them. If such xand y do not exist, print a single number - 1.

Examples

input

2 2 1

output

1 1

input

132 10 35

output

7 5

input

5 18 1000

output

-1

input

48 132 235

output

22 111



1. 计算a/rev(a)的比值，然后以这个比值做排序，然后可以快速算出满足a*b=rev(a)*rev(b)的b的值
2.然后扫描一遍即可

#include <iostream>
#include<vector>
#include<stdio.h>
#include<algorithm>
#define inf 1000000000000
using namespace std;

struct node{
    __int64 x,y,z;
};
vector<node> b,f;
__int64 a[200000];
int tmp[1000],c[200000][3],p[200000];
vector<int>  g[200000];
bool cmp(node a,node b)
{
    return ((a.x*b.y<a.y*b.x));//||(a.x*b.y==a.y*b.x&&a.x<b.x));
}
int dog(int x)
{
    int i,ret=0;
    for(i=1;x!=0;i++)
    {
        tmp[i]=x%10;
        x=x/10;
    }
    for(int j=i-1,k=1;j>=1;j--,k*=10)
        ret+=tmp[j]*k;
    return ret;

}
int main()
{
    int n,mx,my,x1,x2,k1,k2,k3,k4,j,k;
    __int64 w,ans;
    while(scanf("%d%d%I64d",&mx,&my,&w)!=EOF)
    {
        b.clear();
        n=max(mx,my);
        for(int i=1;i<=n;i++)
        {
            a[i]=dog(i);
            node nd;
            nd.x=i;
            nd.y=a[i];
            b.push_back(nd);
        }
        sort(b.begin(),b.end(),cmp);
        f.push_back(b[0]);
        f[0].z=1;
        int t1=1;
        k3=n-1;
        for(int i=0;i<=n;i++)
        {
            g[i].clear();
            p[i]=0;
        }
        for(int i=0;i<n;i=k2+1)
        {
            k1=i;
            for(j=i;j<n;j++)
                if (b[j].x*b[k1].y!=b[j].y*b[k1].x) break;
            k2=j-1;
            for(j=k3;j>=0;j--)
                if (b[i].x*b[j].x<=b[i].y*b[j].y) break;
            k3=j;
            if (k3<0) continue;
            if(b[i].x*b[j].x!=b[i].y*b[j].y) continue;
            for(k=j;k>=0;k--)
                if (b[j].x*b[k].y!=b[j].y*b[k].x) break;
            k4=k+1;
            for(j=k1;j<=k2;j++)
                for(k=k4;k<=k3;k++)
                    g[b[j].x].push_back(b[k].x);
            i=k2+1;
        }
        ans=inf;
        int pt=my,tot=0;
        for(int i=1;i<=mx;i++)
        {
            for(j=0;j<g[i].size();j++)
            {
                if (g[i][j]>pt) continue;
                p[g[i][j]]++;
                tot++;
            }
            if (tot<w) continue;
            if (ans>(__int64)(i)*(__int64)(pt))
            {
                ans=(__int64)(i)*(__int64)(pt);
                x1=i;
                x2=pt;
            }
            for(j=pt;j>=1;j--)
            {
                if (tot-p[j]>=w)
                {
                    tot-=p[j];
                    if (ans>(__int64)(i)*(__int64)(j-1))
                    {
                        ans=(__int64)(i)*(__int64)(j-1);
                        x1=i;
                        x2=j-1;
                    }
                }
                else
                {
                    pt=j;
                    break;
                }
            }
        }
        if (ans!=inf) printf("%d %d\n",x1,x2);
        else
            printf("-1\n");

    }
    return 0;
}

转载于:https://www.cnblogs.com/oi111/p/5260177.html