POJ 3971 Scales

Description

You are given scales for weighing loads. On the left side lies a single stone of known weight W<2  ^N. You own a set of N different weights, weighing 1, 2, 4 ... 2  ^(N-1) units of mass respectively. Determine how many possible ways there are of placing some weights on the sides of the scales; so as to balance them (put them in a state of equilibrium). Output this value modulo a small integer D.

Input

The input begins with an integer t, the number of test cases. Then t test cases follow. For each test case, the first line contains three integers: N L D, where N denotes the number of weights at your disposal, L is the length of the binary representation of number W, and D is the modulus (1<=L<=N<=1000000, 2<=D<=100). The second line contains the value of W, encoded in the binary system as a sequence of exactly L characters 0 or 1 without separating spaces.

Output

For each test case, output a single line containing one integer - the calculated number of possible weight placements, modulo D.

Sample Input

2 6 4 6 1000 6 6 100 100110

Sample Output

3

5

动规，考虑是否进位的问题。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn=1000005;
int n,l,d,f[maxn][2],s[maxn];
char c[maxn];

int main()
{
  int T;
  scanf("%d",&T);
  while (T--)
  {
    memset(f,0,sizeof(f));
    memset(s,0,sizeof(s));
    scanf("%d%d%d",&n,&l,&d);
    scanf("%s",c);
    for (int i=l-1;i>=0;i--) s[l-i-1]=c[i]-'0';
    f[0][0]=1;  f[0][1]=s[0];

    for (int i=1;i<n;i++)
    {
      if (s[i]==0)
      {
        f[i][0]=f[i-1][0]+f[i-1][1];
        f[i][1]=f[i-1][1];
      }
      else 
      {
        f[i][0]=f[i-1][0];
        f[i][1]=f[i-1][0]+f[i-1][1];
      }
      f[i][0]%=d;
      f[i][1]%=d;
    }
    printf("%d\n",(f[n-1][0]+d)%d);
  }
  return 0;
}