B. Pasha and Phone
Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits.
Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a1, a2, ..., an / k and b1, b2, ..., bn / k. Let's split the phone number into blocks of length k. The first block will be formed by digits from the phone number that are on positions 1, 2,..., k, the second block will be formed by digits from the phone number that are on positions k + 1, k + 2, ..., 2·k and so on. Pasha considers a phone number good, if the i-th block doesn't start from the digit bi and is divisible by ai if represented as an integer.
To represent the block of length k as an integer, let's write it out as a sequence c1, c2,...,ck. Then the integer is calculated as the result of the expression c1·10k - 1 + c2·10k - 2 + ... + ck.
Pasha asks you to calculate the number of good phone numbers of length n, for the given k, ai and bi. As this number can be too big, print it modulo 109 + 7.
Input
The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ min(n, 9)) — the length of all phone numbers and the length of each block, respectively. It is guaranteed that n is divisible by k.
The second line of the input contains n / k space-separated positive integers — sequence a1, a2, ..., an / k (1 ≤ ai < 10k).
The third line of the input contains n / k space-separated positive integers — sequence b1, b2, ..., bn / k (0 ≤ bi ≤ 9).
Output
Print a single integer — the number of good phone numbers of length n modulo 109 + 7.
Sample test(s)
input
6 2
38 56 49
7 3 4
output
8
input
8 2
1 22 3 44
5 4 3 2
output
32400
Note
In the first test sample good phone numbers are: 000000, 000098, 005600, 005698, 380000, 380098, 385600, 385698.
题意：将一个长度为n的phone number，分成长度为k的n/k给块，如果任意一个块(块i)中的数字x的开头数字不是b[i], 且x可以整除a[i]，那么就称这个phone number是 good number。问这样的good number 有多少个！
思路：容斥原理。 块 i : tot = 数字长度为k且可以整除a[i]的个数；
bi_tot = 数字长度为k且开头数字为b[i]且可以整除a[i]的个数 - 数字长度为k且开头数字为(b[i]-1)且可以整除a[i]的个数
那么符合要求的数字个数 = tot - bi_tot;(b[i]>0)
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define N 1000005
#define MOD 1000000007
using namespace std;
typedef __int64 LL;
int a[N];
int b[N];
int f[];
void init(){
f[] = ;
for(LL i=; i<; ++i)
f[i] = f[i-] * ;
} int main(){
init();
int n, k;
scanf("%d%d", &n, &k);
int m = n/k;
for(int i=; i<=m; ++i)
scanf("%d", &a[i]);
for(int i=; i<=m; ++i)
scanf("%d", &b[i]);
LL ans = ;
for(int i=; i<=m; ++i){
LL tot = (f[k]-)/a[i] + ;
LL bi_tot0 = (f[k-]-)/a[i] + ;//block的开头是0, 即b[i]==0
LL bi_tot = (f[k-]*(b[i]+)-)/a[i] - (f[k-]*b[i]-)/a[i];//block的开头不是0
if(b[i] == )
ans *= tot-bi_tot0;
else
ans *= tot-bi_tot;
ans %= MOD;
}
printf("%I64d\n", ans);
return ;
}
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2^wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2^a1, ..., 2^ak if and only if there exists a non-negative integer x such that 2^a1 + 2^a2 + ... + 2^ak = 2^x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Print the minimum number of steps in a single line.
5
1 1 2 3 3
2
4
0 1 2 3
4
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
题意：给定n个数w1, w2, w3,.......wn, 然后从这个n数中找出这样的一个子序列a1,a2,a3....ak 使得 2^a1+2^a2.....+2^ak = 2^x, 然后可以删除这个序列，
那么最少经过几步可以全部将这n个数删除！
思路：其实就是 二进制数 相加的过程，n个数对应n个二进制数，从最低位到最高位相加，得到最后的二进制数中 1 的个数就是答案！
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
map<int, int, less<int> >mp;
int main(){
int n;
scanf("%d", &n);
while(n--){
int x;
scanf("%d", &x);
mp[x]++;
}
int ans = ;
for(map<int,int>::iterator it = mp.begin(); it!=mp.end(); ++it){
if(it->second>)
mp[it->first + ] += it->second / ;
if(it->second% != ) ++ans;
}
printf("%d\n", ans);
return ;
}