1073. Square Country
Time limit: 1.0 second
Memory limit: 64 MB
There live square people in a square country. Everything in this country is square also. Thus, the Square Parliament has passed a law about a land. According to the law each citizen of the country has a right to buy land. A land is sold in squares, surely. Moreover, a length of a square side must be a positive integer amount of meters. Buying a square of land with a side a one pays a2 quadrics (a local currency) and gets a square certificate of a landowner.
One citizen of the country has decided to invest all of his N quadrics into the land. He can, surely, do it, buying square pieces 1 × 1 meters. At the same time the citizen has requested to minimize an amount of pieces he buys: "It will be easier for me to pay taxes," — he has said. He has bought the land successfully.
Your task is to find out a number of certificates he has gotten.
Input
The only line contains a positive integer N ≤ 60 000 , that is a number of quadrics that the citizen has invested.
Output
The only line contains a number of certificates that he has gotten.
Sample
| input | output |
|---|---|
344 |
3 |
Problem Author: Stanislav Vasilyev
Problem Source: Ural State Univerisity Personal Contest Online February‘2001 Students Session
Tags: dynamic programming (hide tags for unsolved problems)
Difficulty: 161
题意：给出x，问要多少个完全平方数相加才能得到x。
分析：dp可以解决这个问题。
dp[i]表示i最少要多少个来组成、
dp[i] = min(d[i - j*j] + 1)
但是这题有更简单的做法。
根据定理，仍和正整数可以有四个完全平方数组成。
所以说最多四个数。
这样直接暴力即可。
当然，我是弱智，做的时候没有想到。
1 /**
2 Create By yzx - stupidboy
3 */
4 #include <cstdio>
5 #include <cstring>
6 #include <cstdlib>
7 #include <cmath>
8 #include <deque>
9 #include <vector>
10 #include <queue>
11 #include <iostream>
12 #include <algorithm>
13 #include <map>
14 #include <set>
15 #include <ctime>
16 #include <iomanip>
17 using namespace std;
18 typedef long long LL;
19 typedef double DB;
20 #define MIT (2147483647)
21 #define INF (1000000001)
22 #define MLL (1000000000000000001LL)
23 #define sz(x) ((int) (x).size())
24 #define clr(x, y) memset(x, y, sizeof(x))
25 #define puf push_front
26 #define pub push_back
27 #define pof pop_front
28 #define pob pop_back
29 #define ft first
30 #define sd second
31 #define mk make_pair
32
33 inline int Getint()
34 {
35 int Ret = 0;
36 char Ch = ‘ ‘;
37 bool Flag = 0;
38 while(!(Ch >= ‘0‘ && Ch <= ‘9‘))
39 {
40 if(Ch == ‘-‘) Flag ^= 1;
41 Ch = getchar();
42 }
43 while(Ch >= ‘0‘ && Ch <= ‘9‘)
44 {
45 Ret = Ret * 10 + Ch - ‘0‘;
46 Ch = getchar();
47 }
48 return Flag ? -Ret : Ret;
49 }
50
51 const int N = 60010;
52 int n;
53 int dp[N];
54
55 inline void Input()
56 {
57 cin >> n;
58 }
59
60 inline void Solve()
61 {
62 dp[0] = 0, dp[1] = 1;
63 for(int i = 2; i <= n; i++)
64 {
65 dp[i] = INF;
66 for(int j = 1; j <= i / j; j++)
67 if(dp[i] > dp[i - j * j] + 1)
68 dp[i] = dp[i - j * j] + 1;
69 }
70 cout << dp[n] << endl;
71 }
72
73 int main()
74 {
75 freopen("e.in", "r", stdin);
76 Input();
77 Solve();
78 return 0;
79 }