Ural_1073. Square Country(DP)
思路:dp,转移方程 dp[i] = min(dp[i-1] + 1, dp[i - j*j] + 1);这个转移方程很好理解,类似0-1背包,买j*j的这块地还是不买。跟POJ_3267 The Cow Lexicon很像。
ps:上午wa了好几次,下午昨晚福州的网赛又拿过来这个题,不知道改了哪地方,没积分钟就ac了。看来是老天可怜我网赛被虐。。。T_T
My Code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <math.h>
using namespace std;
const int N = 60005;
int dp[N];
int main(){
//freopen("data.in", "r", stdin);
int n, i, j;
while(cin >> n){
memset(dp, 0, sizeof(dp));
for(i = 1; i <= n; i++){
dp[i] = dp[i-1]+1;
for(j = 1; j <= (int)sqrt((double)n); j++){
if(i >= j*j){
dp[i] = min(dp[i], dp[i - j*j] + 1);
}
}
}
cout << dp[n] << endl;
}
return 0;
}