【HDOJ】4345 Permutation

即求P1^n1+P2^n2 + ... + Pk^nk <= n，其中Pk为素数的所有可能组合。
思路是DP。1~1000的素数就不到200个。
dp[i][j]表示上式和不超过且当前最小素数为P[j]的所有可能情况。注意dp[i][0]+1即为所求。

  1 /* 4345 */
  2 #include <iostream>
  3 #include <sstream>
  4 #include <string>
  5 #include <map>
  6 #include <queue>
  7 #include <set>
  8 #include <stack>
  9 #include <vector>
 10 #include <deque>
 11 #include <algorithm>
 12 #include <cstdio>
 13 #include <cmath>
 14 #include <ctime>
 15 #include <cstring>
 16 #include <climits>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <functional>
 20 #include <iterator>
 21 #include <iomanip>
 22 using namespace std;
 23 //#pragma comment(linker,"/STACK:102400000,1024000")
 24 
 25 #define sti                set<int>
 26 #define stpii            set<pair<int, int> >
 27 #define mpii            map<int,int>
 28 #define vi                vector<int>
 29 #define pii                pair<int,int>
 30 #define vpii            vector<pair<int,int> >
 31 #define rep(i, a, n)     for (int i=a;i<n;++i)
 32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 33 #define clr                clear
 34 #define pb                 push_back
 35 #define mp                 make_pair
 36 #define fir                first
 37 #define sec                second
 38 #define all(x)             (x).begin(),(x).end()
 39 #define SZ(x)             ((int)(x).size())
 40 #define lson            l, mid, rt<<1
 41 #define rson            mid+1, r, rt<<1|1
 42 
 43 const int maxn = 1005;
 44 bool isPrime[maxn];
 45 int P[maxn], pn = 0;
 46 __int64 dp[maxn][205];
 47 __int64 ans[maxn];
 48 
 49 void init() {
 50     memset(isPrime, true, sizeof(isPrime));
 51     rep(i, 2, maxn) {
 52         if (isPrime[i]) {
 53             P[pn++] = i;
 54             for (int j=i*i; j<maxn; j+=i)
 55                 isPrime[j] = false;
 56         }
 57     }
 58     
 59     #ifndef ONLINE_JUDGE
 60         printf("pn = %d\n", pn);
 61     #endif
 62     ans[1] = 1;
 63     rep(i, 2, 1001) {
 64         rep(j, 0, pn) {
 65             if (P[j] > i)
 66                 break;
 67             for (int k=P[j]; k<=i; k*=P[j]) {
 68                 dp[i][j] += dp[i-k][j+1] + 1;
 69             }
 70         }
 71         per(j, 0, 200)
 72             dp[i][j] += dp[i][j+1];
 73         ans[i] = dp[i][0] + 1;
 74     }
 75     
 76     #ifndef ONLINE_JUDGE
 77         rep(i, 1, 1001)
 78             printf("%d: %I64d\n", i, ans[i]);
 79     #endif
 80 }
 81 
 82 int main() {
 83     ios::sync_with_stdio(false);
 84     #ifndef ONLINE_JUDGE
 85         freopen("data.in", "r", stdin);
 86         freopen("data.out", "w", stdout);
 87     #endif
 88     
 89     int n;
 90     
 91     init();
 92     
 93     while (scanf("%d", &n)!=EOF)
 94         printf("%I64d\n", ans[n]);
 95     
 96     #ifndef ONLINE_JUDGE
 97         printf("time = %d.\n", (int)clock());
 98     #endif
 99     
100     return 0;
101 }

 

posted on 2016-02-13 22:02  Bombe  阅读(174)  评论(0)  编辑  收藏  举报

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