Description
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
input | output |
---|---|
344 |
3 |
大意:你现在有n块土地,要把它分成几块正方形,问最少的正方形数目,原本以为可以贪心做,倒着过来,不过12(4+4+4 = 3) 9+1+1+1 = 4这组数据有问题
动态转移方程 dp[i] = min(dp[i],dp[i-j*j)+1)
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int main() { int dp[60010]; int n; while(~scanf("%d",&n)){ memset(dp,0,sizeof(dp)); for(int i = 1; i <= n ;i++){ dp[i] = dp[i-1] + 1; for(int j = 1; j <= sqrt((double)n+0.5); j++) if(j*j <= i) dp[i] = min(dp[i],dp[i-j*j] + 1); } printf("%d\n",dp[n]); } return 0; }