HDU 4658 Integer Partition (2013多校6 1004题)
Integer Partition
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22 Accepted Submission(s): 15
Problem Description
Given n, k, calculate the number of different (unordered) partitions of n such that no part is repeated k or more times.
Input
First line, number of test cases, T.
Following are T lines. Each line contains two numbers, n and k.
1<=n,k,T<=105
Following are T lines. Each line contains two numbers, n and k.
1<=n,k,T<=105
Output
T lines, each line contains answer to the responding test case.
Since the numbers can be very large, you should output them modulo 109+7.
Since the numbers can be very large, you should output them modulo 109+7.
Sample Input
4
4 2
4 3
4 4
4 5
Sample Output
2
4
4
5
Source
Recommend
zhuyuanchen520
跟上次多校求数的划分很类似。
所谓的五边形数定理还没有搞懂。
先贴个代码先,胡搞弄过去的
1 /* 2 * Author: kuangbin 3 * Created Time: 2013/8/8 11:53:35 4 * File Name: 1004.cpp 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <cstring> 10 #include <cmath> 11 #include <algorithm> 12 #include <string> 13 #include <vector> 14 #include <stack> 15 #include <queue> 16 #include <set> 17 #include <time.h> 18 using namespace std; 19 const int MOD = 1e9+7; 20 int dp[100010]; 21 void init() 22 { 23 memset(dp,0,sizeof(dp)); 24 dp[0] = 1; 25 for(int i = 1;i <= 100000;i++) 26 { 27 for(int j = 1, r = 1; i - (3 * j * j - j) / 2 >= 0; j++, r *= -1) 28 { 29 dp[i] += dp[i -(3 * j * j - j) / 2] * r; 30 dp[i] %= MOD; 31 dp[i] = (dp[i]+MOD)%MOD; 32 if( i - (3 * j * j + j) / 2 >= 0 ) 33 { 34 dp[i] += dp[i - (3 * j * j + j) / 2] * r; 35 dp[i] %= MOD; 36 dp[i] = (dp[i]+MOD)%MOD; 37 } 38 39 } 40 41 42 43 } 44 } 45 46 int solve(int n,int k) 47 { 48 int ans = dp[n]; 49 for(int j = 1, r = -1; n - k*(3 * j * j - j) / 2 >= 0; j++, r *= -1) 50 { 51 ans += dp[n -k*(3 * j * j - j) / 2] * r; 52 ans %= MOD; 53 ans = (ans+MOD)%MOD; 54 if( n - k*(3 * j * j + j) / 2 >= 0 ) 55 { 56 ans += dp[n - k*(3 * j * j + j) / 2] * r; 57 ans %= MOD; 58 ans = (ans+MOD)%MOD; 59 } 60 61 } 62 return ans; 63 } 64 65 int main() 66 { 67 init(); 68 int T; 69 int n,k; 70 scanf("%d",&T); 71 while(T--) 72 { 73 scanf("%d%d",&n,&k); 74 printf("%d\n",solve(n,k)); 75 } 76 return 0; 77 }
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