枚举--hdu5936 difference

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

 

//即求0 <= x <= 1e9时有多少个y满足x + y == f(y,k)

//1 <= k <= 9,f(y,k)min = f(y,1) < ..< f(y,i) < ... < f(y,k)max = f(y,9) f(y,k)关于k是递增函数

//y有1位时,可以知道y == f(y,1) < f(y,i) 此时x取0即可

//y有2位时，f(y,k)min < y < f(y,k)max,存在正整数x使得等式x + y == f(y,k)成立

//假设y有m位时不存在正整数x使等式成立，f(y,k)max < 10^(m - 1) <= y <= x + y

//即f(y,k)max <= m * 9^9 < 10 ^(m - 1)成立，m = 9时还不成立，m >= 10时不等式恒成立，不等式成立即y有10位时，不存在正整数x使得x + y == f(y,k)

//所以y最多有10位数字

//啊啊啊啊啊啊x==0的时候，y = 0也满足，要减掉1

#include <iostream>

#include <cstdio>

#include <cmath>

#include <map>

using namespacestd;

typedef longlong ll;

ll ak[10][10];//a的k次方

map<ll,ll> cnt[10];

void init_ak()

{

    for (int i =0; i < 10; i ++) {

        for (int j =1; j < 10; j ++) {

            ak[i][j] =pow(i, j);

        }

    }

}

ll f(int y,int k)

{

    ll sum =0;

    while (y) {

        sum += ak[y %10][k];

        y /= 10;

    }

    return sum;

}

void init_y()

{

    ll sum =0;

    for (int k =1; k <= 9; k ++) {

        for (int y =0; y < 1e5; y ++) {

            sum = f(y, k);

            cnt[k][sum -y * 1e5] ++;

        }

    }

}

int main()

{

    init_ak();

    init_y();

    int T;

    cin >> T;

    for (int t =1; t <= T; t ++) {

        int x,k;

        scanf("%d%d",&x,&k);

        ll ans =0,key = 0;

        for (int i =0; i < 1e5; i ++) {

            key = x - f(i,k) + i;//枚举y2

            if(cnt[k].find(key) !=cnt[k].end()) {ans +=cnt[k][key];}

        }

        if(x ==0) ans --;

        printf("Case #%d: %lld\n",t,ans);

    }

    return0;

}