poj 2229 Sumsets 完全背包求方案总数
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Sumsets
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
Source
#include<stdio.h> #include<string.h> #define ll long long #define mod 1000000007 #define esp 0.00000000001 const int N=1e5+10,M=1e6+10,inf=1e9; int a[22]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288}; int dp[M]; int main() { int x,y,z,i,t; memset(dp,0,sizeof(dp)); dp[0]=1; for(i=0;i<20;i++) { for(t=a[i];t<=1000000;t++) dp[t]+=dp[t-a[i]],dp[t]%=inf; } while(~scanf("%d",&x)) printf("%d\n",dp[x]); return 0; }
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