Sumsets
http://acm.hdu.edu.cn/showproblem.php?pid=2709
Problem Description
Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7:
1) 1+1+1+1+1+1+1
2) 1+1+1+1+1+2
3) 1+1+1+2+2
4) 1+1+1+4
5) 1+2+2+2
6) 1+2+4
Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000).
1) 1+1+1+1+1+1+1
2) 1+1+1+1+1+2
3) 1+1+1+2+2
4) 1+1+1+4
5) 1+2+2+2
6) 1+2+4
Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000).
Input
A single line with a single integer, N.
Output
The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).
Sample Input
7
Sample Output
6
解题代码:
/* 设way[n]为和为 n 的种类数; 根据题目可知,加数为2的N次方,即 n 为奇数时等于它前一个数 n-1 的种类数 way[n-1] ,若 n 为偶数时分加数中有无 1 讨论,即关键是对 n 为偶数时进行讨论: 1.如果所求的n为奇数,那么所求的分解结果中必含有1,因此,直接将n-1的分拆结果中添加一个1即可 为way[n-1] 2.如果所求的n为偶数,那么n的分解结果分两种情况 (1).如果加数里含1,则一定至少有两个1,即对n-2的每一个加数式后面 +1+1,总类数为way[n-2]; (2).不含有1 那么,分解因子的都是偶数,将每个分解的因子都除以2,刚好是n/2的分解结果,并且可以与之一一对应,这种情况有 s[n/2] 所以,状态转移方程为: 如果i为奇数 way[n] = way[n-1] 如果i为偶数 way[n] = way[n-2]+way[n/2]; */ #include <stdio.h> #include <math.h> #include <string.h> #ifdef WINDOWS #define LL __int64 #define LLd "%I64d" #else #define LL long long #define LLD "%lld" #endif const int max_n = 1e6+2; const LL MOD = 1e9; LL way[max_n]; void deal() { way[0] = way[1] = 1; for (int i = 2; i < max_n; i ++) { if (!(i&1)) { way[i] = way[i-2] + way[i/2]; way[i] %= MOD; } else { way[i] = way[i-1]; way[i] %= MOD; } } } int main () { int n; deal(); while (~scanf ("%d", &n)) { printf (LLD" ", way[n]); } return 0; }