SRM 572 D2L3：DistinctRemainders，dp，math

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题目：http://community.topcoder.com/stat?c=problem_statement&pm=12384&rd=15492

参考：http://apps.topcoder.com/wiki/display/tc/SRM+572

把数学部分搞定之后dp比较简单，关键是找到所取数的模集与最终序列种数的关系。

代码：

#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <iostream>#include <sstream>#include <iomanip>#include <bitset>#include <string>#include <vector>#include <stack>#include <deque>#include <queue>#include <set>#include <map>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <cstring>#include <ctime>#include <climits>using namespace std;#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)typedef pair<int, int> pii;typedef long long llong;typedef pair<llong, llong> pll;#define mkp make_pair/*************** Program Begin **********************/const int MOD = 1e9 + 7;long long dp[51][51][2501];class DistinctRemainders {public:long long N;int M;int rec(int cur, int K, int q){long long & res = dp[cur][K][q];if (res != -1) {return res;}if (M == cur) {// base casesif (q % M == N % M) {long long Q = (N - q) / M;res = K;// C(Q + K - 1, K - 1) * K!for (int i = 0; i < K - 1; i++) {res *= ( (Q + K - 1 - i) % MOD );res %= MOD;}return res;} else {res = 0;return res;}}res = 0;// addres += rec(cur + 1, K + 1, q + cur);res %= MOD;// ignoreres += rec(cur + 1, K, q);res %= MOD;return (int)res;}int howMany(long long N, int M) {int res = 0;this->N = N;this->M = M;memset(dp, -1, sizeof(dp));res = rec(0, 0, 0);return res;}};/************** Program End ************************/

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