LA 6801 Sequence (DP)

LA 6801 Sequence (DP)

6801-Sequence

Question:

Given the initial status of n switches (0/1), perform k operations. You can select any one of them and reverse the status (0-1, 1-0 ). Question: The number of methods that make the final state all 0% 1000000007.

Solution:

Dynamic Planning. Dp [I] [j] indicates the number of j 1 solutions after the I operation.

State transition equation: dp [I] [j + 1] = dp [I-1] [j] * (n-j)

Dp [I] [J-1] = dp [I-1] [j] * j

Reference code:

// Author: Yuan Zhu#include 
 
  #include 
  
   #include 
   
    #define ll long long#define mod 1000000007using namespace std;int t, n, k;int a[1100];ll dp[1100][1100];int main() {    scanf("%d", &t);    for (int ca = 1; ca <= t; ca++) {        memset(dp, 0, sizeof dp);        scanf("%d%d", &n, &k);        int one = 0;        for (int i = 1; i <= n; i++) {            scanf("%d", a + i);            if(a[i] == 1) one++;        }        dp[0][one] = 1;        for (int i = 0; i < k; i++) {            for (int j = 0; j <= n; j++) {                if (j) dp[i + 1][j - 1] = (dp[i + 1][j - 1] + dp[i][j] * j) % mod;                if (j < n) dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j] * (n - j)) % mod;            }        }        printf("Case #%d: %lld\n", ca, (dp[k][0] % mod + mod) % mod);    }    return 0;}