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题目链接:点击打开链接
思路:DP + 组合数。 用d[i][j]表示前第i种颜色的石头, 已经用了j个的方法数, 每次枚举第i种石头放多少个, 假设放k个, 那么相当于从j个位置中选k个, 预处理组合数就行了。
细节参见代码:
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) using namespace std; typedef long long ll; typedef long double ld; const ld eps = 1e-9, PI = 3.1415926535897932384626433832795; const int mod = 1000000000 + 7; const int INF = 0x3f3f3f3f; // & 0x7FFFFFFF const int seed = 131; const ll INF64 = ll(1e18); const int maxn = 100 + 10; int T,n,m,kase = 0,a[maxn]; ll d[2][maxn*maxn]; ll C[maxn*maxn][maxn]; inline void add(ll &a, ll b) { a += b; if(a >= mod) a -= mod; } void init() { for(int i = 0; i <= 10000; i++) { C[i][0] = 1; for(int j = 1; j <= min(i, 100); j++) { C[i][j] = C[i-1][j-1] ; add(C[i][j], C[i-1][j]); } } } int main() { init(); while(~scanf("%d",&n)) { for(int i = 1; i <= n; i++) scanf("%d",&a[i]); int u = 1; memset(d[u], 0, sizeof(d[u])); d[u][0] = 1; int maxlen = 0; for(int i = 1; i <= n; i++) { maxlen += a[i]; memset(d[u^1], 0, sizeof(d[u^1])); for(int j = 0; j <= maxlen; j++) { for(int k = 0; k <= a[i]; k++) { if(k > j || d[u][j-k] == 0) continue; ll cur = C[j][k] * d[u][j-k]; if(cur >= mod) cur %= mod; add(d[u^1][j] , cur); } } u ^= 1; } ll ans = 0; for(int i = 1; i <= maxlen; i++) { add(ans , d[u][i]); } printf("Case %d: %I64d\n",++kase, ans); } return 0; }
HDU 4248 A Famous Stone Collector(DP + 组合数)
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原文地址:http://blog.csdn.net/weizhuwyzc000/article/details/51204360