标签：com http blog class div code c log t sp string
http://acm.nyist.net/JudgeOnline/problem.php?pid=708
状态转移方程的思路:对于一个数N,可以是N - 1的状态+1 得到,另外,也可以是(n / 2) * (1 + 1)得到,同理对于任意的奇数p,都有如果n可以整除p,都有f(n / p) + f(p)
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51 |
#include<stdio.h>int
dp[10010];bool
isprime[101];int
prime[101];int
total;void
sol(){ int
i; for(i = 4; i < 10010; i++) { int
min = dp[i - 1] + 1; int
j; for(j = 0; prime[j] < i && j < total; j ++) { if(i % prime[j] == 0) { min = min < (dp[i / prime[j]] + dp[prime[j]]) ? min : dp[i / prime[j]] + dp[prime[j]]; } } dp[i] = min; }}int
main(){ int
n; dp[1] = 1; dp[2] = 2; dp[3] = 3; int
i, j; for(i = 2; i * i < 100; i++) { if(isprime[i] == 0) { for(j = i + i; j < 100; j+= i) { isprime[j] = 1; } } } for(i = 2; i < 100; i++) { if(isprime[i] == 0) prime[total ++] = i; } sol(); while(scanf("%d",&n) != EOF) { printf("%d\n", dp[n]); } return
0;} |
nyoj 708 ones 动态规划,布布扣,bubuko.com
标签：com http blog class div code c log t sp string
原文地址：http://www.cnblogs.com/z-y-p/p/3698952.html
