HDU5936-Difference

Difference

                                                                              Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                           Total Submission(s): 850    Accepted Submission(s): 236

Problem Description

Little Ruins is playing a number game, first he chooses two positive integers y and K and calculates f(y,K), here

f(y,K)=∑z in every digits of yzK(f(233,2)=22+32+32=22)

then he gets the result

x=f(y,K)−y

As Ruins is forgetful, a few seconds later, he only remembers K, x and forgets y. please help him find how many y satisfy x=f(y,K)−y.

 

Input

First line contains an integer T, which indicates the number of test cases.

Every test case contains one line with two integers x, K.

Limits
1≤T≤100
0≤x≤109
1≤K≤9

 

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the result.

 

Sample Input

22 23 2

 

Sample Output

Case #1: 1Case #2: 2

 

Source

2016年中国大学生程序设计竞赛（杭州）

 

题意：给你数字x和k，问有多少个数字满足它每位数字k次方的和等于x和它的和

解题思路：可以用折半搜索的方法，先预处理出0~100000所有数每位数字k次方的和，然后先枚举前5位，记下相应的值，二分查找后5位，与前面的匹配看是否能够凑成x，最后统计答案。（本题x等于0时，因为题目y要求大于0，所以答案要建一）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <set>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;struct node{    LL sum1,sum2;} a[10][100005];LL mypow(int x,int y){    LL ans=1;    for(int i=1; i<=y; i++) ans*=1LL*x;    return ans;}bool cmp(const node a,const node b){    return a.sum2<b.sum2;}int main(){    memset(a,0,sizeof a);    for(int i=0; i<100000; i++)    {        int k=i;        for(int j=1; j<=9; j++) a[j][i].sum1-=1LL*100000*i,a[j][i].sum2-=1LL*i;        while(k>0)        {            LL kk=k%10;            LL x=kk;            for(int j=1; j<=9; j++)            {                a[j][i].sum1+=x;                a[j][i].sum2+=x;                x*=kk;            }            k/=10;        }    }    for(int i=1; i<=9; i++) sort(a[i],a[i]+100000,cmp);    int kkk,t,cas=0;    LL x;    scanf("%d",&t);    while(t--)    {        printf("Case #%d: ",++cas);        scanf("%lld%d",&x,&kkk);        LL cnt=0;        for(int i=0; i<100000; i++)        {            LL xx=x-a[kkk][i].sum1;            int l=0,r=99999,ans1=-1,ans2=-1;            while(l<=r)            {                int mid=(l+r)>>1;                if(a[kkk][mid].sum2>xx) r=mid-1;                else if(a[kkk][mid].sum2<xx) l=mid+1;                else ans1=mid,r=mid-1;            }            if(ans1==-1) continue;            l=0,r=99999;            while(l<=r)            {                int mid=(l+r)>>1;                if(a[kkk][mid].sum2>xx) r=mid-1;                else if(a[kkk][mid].sum2<xx) l=mid+1;                else ans2=mid,l=mid+1;            }            cnt+=1LL*(ans2-ans1+1);        }        if(x==0) cnt--;        printf("%lld\n",cnt);    }    return 0;}