摘要:
原题链接:http://acm.timus.ru/problem.aspx?space=1&num=1073分析:dp,dp[i]表示钱为i且恰好用完时能买的最少土地数,易知dp[i]=min(i,dp[i-j*j]+1)(1#include#include#include#include#include#include#include#define ll long long#define maxn 60005using namespace std;int dp[maxn];void solve(){ memset(dp,0,sizeof(dp)); dp[0]=0;dp[1]... 阅读全文