hihoCoder挑战赛5 C 与链

有两种DP搞法，不过其实本质上是一样的。。。

一种是按照题解上说的记录当前到i位，进位为j的种类数，转移的时候直接枚举在这一位上面放多少个1就好了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>

using namespace std;

typedef long long LL;
const LL mod = 1e9 + 9;

int bits[64], k, n;
LL f[64][10000 + 10];

void getbits(int num) {
	memset(bits, 0, sizeof(bits));
	for(int i = 0; i <= 30; i++) bits[i] = (bool)(num & (1 << i));
}

LL dfs(int now, int over, LL nownum) {
	if(now == 30) return over == 0;
	if(nownum > n) return 0;
	if(f[now][over] != -1) return f[now][over];
	LL ret = 0;
	for(int i = 0; i <= k; i++) {
		if(nownum + (i << now) > n) break;
		if((over + i) % 2 == bits[now]) {
			ret += dfs(now + 1, (over + i - bits[now]) >> 1, 
					nownum + (i << (now)));
			ret %= mod;
		}
	}
	if(f[now][over] == -1) f[now][over] = ret % mod;
	return ret % mod;
}

int main() {
	int T; scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &k, &n);
		memset(f, -1, sizeof(f));
		getbits(n);
		cout << dfs(0, 0, 0) << endl;
	}
	return 0;
}

　　

另外一种是记录当前位i构成和为j的种类数，转移的时候也是枚举在当前为放多少个1.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>

using namespace std;
const int maxn = 1e4 + 10;
typedef long long LL;
const LL mod = 1e9 + 9;
LL f[30][maxn];

int main() {
	int T; scanf("%d", &T);
	while(T--) {
		int n, k; scanf("%d%d", &k, &n);
		memset(f, 0, sizeof(f));
		for(int i = 0; i <= k && i <= n; i++) f[0][i] = 1;
		for(int i = 0; i < 16; i++) {
			for(int j = 0; j <= n; j++) if(f[i][j]) {
				for(int t = 0; t <= k; t++) {
					LL nxt = ((LL)t << (i + 1)) + j;
					if(nxt > n) break;
					f[i + 1][nxt] = (f[i + 1][nxt] + f[i][j]) % mod;
				} 
			}
		}
		cout << f[16][n] << endl;
	}
	return 0;
}