Codeforces Round #287 (Div. 2) D. The Maths Lecture （数位dp）

题意 ： 位数为n的数中有几个数的后缀能被k整除

数位dp

int  n, k;LL m;LL dp[1111][111][2], mul[1111], mulm[1111];LL dfs ( int cur, int rest, int e ) {  //当前位数为cur， 余数为rest， e == 1 表示有前导0LL &ans = dp[cur][rest][e];if( ans != -1 ) return ans;ans = 0;if( cur == n+1 ) return ans;int st = 0;if( cur == n ) st = 1;for( int i = st; i < 10; ++i ) {int nxt = ( rest + mul[cur-1] * i ) % k;int f = 0;if( e && i == 0 ) f = 1;if( nxt == 0 && ! f ) {  //当余数为0时，前面n-cur个位的数可以任选int tmp = n - cur - 1;if( tmp < 0 ) ans = ( ans + 1 ) % m;else ans = ( ans + mulm[tmp]*9%m ) % m;}else ans = ( ans + dfs ( cur + 1, nxt, f ) ) % m;}return ans;}int main () {while( ~scanf("%d%d%I64d", &n, &k, &m ) ) {memset( dp, -1, sizeof( dp ) );mulm[0] = mul[0] = 1;for( int i = 1; i <= n; ++i )mul[i] = mul[i-1] * 10 % k, mulm[i] = mulm[i-1] * 10 % m;printf("%I64d\n" , dfs( 1, 0, 1 ) );}}