标签:nta lse 进制 ios i++ cow 两种 std count
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). 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
对不起,是我太菜了,看到题目又没思路,接着参考大佬的博客
首先定义状态:d[i] 表示i的划分方法数
关键是这里的递推关系也就是状态转移方程:
1.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i-1的分拆结果中添加一个1即d[i] = d[i-1]
2.所求的n为偶数,那么n的分解结果分两种情况
综上:d[i] = d[i-1] (i为奇数)
d[i] = d[i-2] + d[i/2] (i为偶数)
最后由于只要输出最后9个数位,别忘记模1000000000
附上AC代码:
#include<iostream> using namespace std; int d[1000005]; int main() { int i,n; d[1]=1; d[2]=2; for(i=3;i<=1000000;i++) { if(i&1) d[i]=d[i-1]; else d[i]=(d[i-2]+d[i/2])%1000000000; } cin>>n; cout<<d[n]<<endl; return 0; }
附:
i&1用于判断是否为奇数数!如果为真,则为奇数,为假则为偶数
解释:&符号代表 按位与,1的二进制最后一位为1,其余为零。如果一个数为奇数,那么最后一位必为1,其余位必为0,所以得出结果为1。如果是偶数的话,最后一位必然为0,其余位与0与运算必为0,所以结果为0,这样就可以起到判断奇数偶数的效果
标签:nta lse 进制 ios i++ cow 两种 std count
原文地址:https://www.cnblogs.com/wizarderror/p/10478573.html