Digit Division
Time limit: 1 s Memory limit: 512 MiB
We are given a sequence of n decimal digits. The sequence needs to be partitioned into one or more contiguous subsequences such that each subsequence, when interpreted as a decimal number, is divisible by a given integer m.
Find the number of different such partitions modulo 109 + 7. When determining if two partitions are different, we only consider the locations of subsequence boundaries rather than the digits themselves, e.g. partitions 2|22 and 22|2 are considered different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 300 000, 1 ≤ m ≤ 1 000 000) – the length of the sequence and the divisor respectively. The second line contains a string consisting of exactly n digits.
Output
Output a single integer – the number of different partitions modulo 109 + 7.
Example
input
4 2
1246
output
4
input
4 7
2015
output
0
//题意: n 位长的十进制数字，在其中可以任意插入分割线，分割后，要使每一段不为空，并且可以整除 m ，合法分割的方案数
//题目是极其简单的，如果前一部分可以整除 m ，那么，这部分乘10的x次方后依然可以整除，然后算出所有可分割的位置后
C(0,all),C(1,all)+...+C(all,all); 这些必然合法
= 2^all
但是，此题如果没想清楚，写代码会进坑，此题是对方案数取模，all 是%m==0的方案数，进了坑半天想不出来，唉，还是太菜啊，一度wa在第三组，真是日狗了
1 # include <cstdio> 2 # include <cstring> 3 # include <cstdlib> 4 # include <iostream> 5 # include <vector> 6 # include <queue> 7 # include <stack> 8 # include <map> 9 # include <bitset> 10 # include <sstream> 11 # include <set> 12 # include <cmath> 13 # include <algorithm> 14 # pragma comment(linker,"/STACK:102400000,102400000") 15 using namespace std; 16 # define LL long long 17 # define pr pair 18 # define mkp make_pair 19 # define lowbit(x) ((x)&(-x)) 20 # define PI acos(-1.0) 21 # define INF 0x3f3f3f3f 22 # define eps 1e-8 23 # define MOD 1000000007 24 25 inline int scan() { 26 int x=0,f=1; char ch=getchar(); 27 while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();} 28 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} 29 return x*f; 30 } 31 inline void Out(int a) { 32 if(a<0) {putchar('-'); a=-a;} 33 if(a>=10) Out(a/10); 34 putchar(a%10+'0'); 35 } 36 # define MX 300050 37 /**************************/ 38 char num[MX]; 39 40 LL qk_mi(LL base,LL x) 41 { 42 LL res = 1; 43 while (x) 44 { 45 if (x%2==1) res = (res*base)%MOD; 46 base=base*base%MOD; 47 x/=2; 48 } 49 return res; 50 } 51 52 int main() 53 { 54 int n,m; 55 while (scanf("%d%d",&n,&m)!=EOF) 56 { 57 scanf("%s",num); 58 LL zuo = 0; 59 LL all = 0; 60 for (int i=0;i<n;i++) 61 { 62 zuo=(zuo*10+(num[i]-'0'))%m; 63 if (zuo%m==0) all++; 64 } 65 all--; 66 if (zuo%m!=0) 67 printf("0 "); 68 else 69 { 70 LL ans = qk_mi(2,all); 71 printf("%lld ",ans); 72 } 73 } 74 return 0; 75 }