51nod 1043 幸运号码(数位dp)

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #define CLR(a,b) memset((a),(b),sizeof((a)))
 5 using namespace std;
 6 
 7 const int inf = 0x3f3f3f3f;
 8 const int mod = 1e9 + 7;
 9 const int N = 1001;
10 int n;
11 long long dp[2][9*N];//i个数和为j的数量
12 int main(){
13     int i, j, k;
14     long long sum, ans;
15     scanf("%d", &n);
16     CLR(dp, 0);
17     for(i = 0; i <= 9; ++i)
18         dp[1][i] = 1;
19     for(i = 2; i <= n; ++i){
20         for(j = 0; j <= 9*i; ++j){
21             sum = 0;
22             for(k = 0; k <= 9; ++k){
23                 if(j >= k)
24                     sum = (sum + dp[(i-1)&1][j-k]) % mod;
25             }
26             dp[i&1][j] = sum;
27         }
28 
29     }
30     ans = 0;
31     for(i = 1; i <= 9*n; ++i){
32         ans = (ans + (dp[n&1][i]-dp[(n-1)&1][i]) *dp[n&1][i]) %mod;
33     }
34     printf("%d\n", ans);
35     return 0;
36 }