hdu 5936 二分好题 2016年中国大学生程序设计竞赛（杭州）

Difference

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 933    Accepted Submission(s): 259

Problem Description

Little Ruins is playing a number game, first he chooses two positive integersy 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', wherex indicates the case number and counts from 1 andy is the result.

 

Sample Input

22 23 2

 

Sample Output

Case #1: 1Case #2: 2

题意：

看上面的式子

题解：

将y这个数分为两半，先处理后一半，然后枚举后一半再二分查找前一半，判断是否可以满足上面的式子

题目卡时间看的很气，数组开 long long  不行，会超时

并且再指数次方的那里需要预处理，并且不能先处理前一半，因为后一半和前一半有乘以100000的一个关系，利用好又可以减少时间

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 100005#define LL long longint a[maxn],sum[20][20],num[maxn];int main(){    for(int i=1;i<10;i++){        sum[i][0]=1;        for(int j=1;j<=9;j++){            sum[i][j]=sum[i][j-1]*i;        }    }    int T;    int cases=1,x,k,temp,pos;    //freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--)    {        LL t;        pos=0;        scanf("%d%d",&x,&k);        memset(num,0,sizeof(num));        for(int i=1;i<100000;i++){            temp=i;            while(temp)            {                num[i]+=sum[temp%10][k];                temp/=10;            }            a[pos++]=num[i]-i;        }        sort(a,a+pos);        LL ans=0;        for(LL i=0;i<100000;i++){            t=num[i]-i*100000;            t=x-t;            temp=lower_bound(a,a+pos,t)-a;            while(a[temp]==t&&temp<pos)                ans++,temp++;        }        printf("Case #%d: %lld\n",cases++,ans);    }    return 0;}