数论（同余+hash）
Description You are given a sequence a[0]a[1] ... a[N-1] of digits and a prime number Q. For each i<=j with a[i] != 0, the subsequence a[i]a[i+1]...a[j] can be read as a decimal representation of a positive integer. Subsequences with leading zeros are not considered. Your task is to count the number of pairs (i, j) such that the corresponding subsequence modulo Q is R. Input The input consists of at most 10 datasets. Each dataset is represented by a line containing four integers N, S, W, Q, and R, separated by spaces, where 1<=N<=10^5, 1<=S<=10^9, 1<=W<=10^9, Q is a prime number less than 10^9 and 0<=R<Q. The sequence a[0]...a[N-1] of length N is generated by the following code.
int g = S;
for (int i=0; i<N; i++) {
a[i] = (g/7) % 10;
if( g%2 == 0 ) { g = (g/2); }
else { g = (g/2) ^ W; }
}
Note: the operators /, %, and ^ are the integer division, the modulo, and the bitwise exclusiveor, respectively. The above code is meant to be a random number generator. The intended solution does not rely on the way how the
sequence is generated. The end of the input is indicated by a line containing five zeros separated by spaces. Output For each dataset, output the answer in a line.
Sample Input
3 32 64 7 0
4 35 89 5 0
5 555 442 3 0
5 777 465 11 0
100000 666 701622763 65537
0 0 0 0 0 0
Sample Output
2
4
6
3
68530
Hint In the first dataset, the sequence is 421. We can find two multiples of Q = 7, namely, 42 and 21. In the second dataset, the sequence is 5052, from which we can find 5, 50, 505, and 5 being the multiples of Q = 5. Notice that we don't count 0 or 05 since they are not a valid representation of positive integers. Also notice that we count 5 twice, because it occurs twice in different positions. In the third and fourth datasets, the sequences are 95073 and 12221, respectively.
题目大意
给出一串十进制的数字，求当中有多少个子串所表示的数字模质数p的值为r
思路
如果一个子串为x[i]x[i+1]...x[j],则模p的值为(x[i]*10^(j-i)+x[i+1]*10^(j-i-1)+...+x[j]*10^0)%p,如果用f[n]表示(10^n)%p, t[n]表示原串的后n位模p的值。那么x[i]x[i+1]...x[j] %p=r等价于t[i]=(t[j-1]+r*10^(j-1))%p=(t[j-1]+r*f[j-1])%p,((a/b)%c=(a%(b*c))，这样以第i位开头的数字中符合要求的个数即为符合该式的j的个数（j>=i)。
能够用map或哈希记录个数。m[x]表示符合(t[j]+r*f[j])%p=x的j的个数。
做法
按题目给的方法生成数字串，预处理求出f[n],再从右向左依次处理出t[n],将答案加上m[t[n]],再更新m[(t[n]+r*f[n])%p]就可以。
说明
当题目中的质数为2或5时须要特殊处理，由于10能够被2或5整除，不能再用10的幂次的形式取余。此时仅仅须要从高位向地位。不断统计最后一位模p为r的子串个数就可以，由于一个数除2或5的余数就等于最后一位模2或5.
数字的首位不能为0，所以要注意仅仅有当前位不为0时才干更新答案。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
const int N=100005;
long long a[N];
long long f[N];
void init(int n,long long s,long long w)
{
long long g=s;
for (int i=1; i<=n; i++)
{
a[i]=(g/7)%10;
if (g%2==0) g=g/2;
else g=(g/2)^w;
}
}
int main()
{
int n,i;
long long s,w,q,r;
long long x,y,ans;
while (scanf("%d%lld%lld%lld%lld",&n,&s,&w,&q,&r)==5)
{
if (n==0 && s==0 && w==0 && q==0 && r==0) break;
init(n,s,w); ans=0;
if (q==2 || q==5)
{
s=0;
for (i=1; i<=n; i++){
if (a[i]!=0) s++;
if (a[i]%q==r) ans+=s;
}
}
if (q!=2 && q!=5)
{
f[n]=1;
for (i=n-1; i>=1; i--) f[i]=(f[i+1]*10)%q;
map<int,int> m;
m[r]=1;
x=0;
for (i=n; i>=1; i--)
{
x=(x+a[i]*f[i])%q;
if (a[i]!=0)ans+=m[x];
m[(r*f[i-1]+x)%q]++;
}
}
printf("%lld\n",ans);
}
return 0;
} 