NO.1
思路:哈希+质数判断。
代码实现:
#include <iostream>
#include<string>
#include<cmath>
using namespace std;bool isprime(int n)
{if(n<2) return false;for(int i=2;i<=sqrt(n);i++){if(n%i==0) return false;}return true;
}
string s;int main() {cin>>s;int sum[26]={0};for(auto ch:s){sum[ch-'a']++;}int maxn=0,minn=1000;for(int i=0;i<26;i++){if(sum[i]){maxn=max(sum[i],maxn);minn=min(sum[i],minn);}}if(isprime(maxn-minn)){cout<<"Lucky Word"<<endl;cout<<maxn-minn<<endl;}else {{cout<<"No Answer"<<endl;cout<<0<<endl;}}return 0;
}
NO.2
思路:先给左端点进行排序,如果该区间的左端点小于前一个区间的右端点那么就返回false,反之返回true。
代码实现:
class Solution {
public:bool hostschedule(vector<vector<int> >& schedule) {sort(schedule.begin(),schedule.end());for(int i=1;i<schedule.size();i++){if(schedule[i][0]<schedule[i-1][1]) return false;}return true;}
};
NO.3
思路: 背包问题:原问题转换成,从 n 个数中选,总和恰好为 sum / 2,能否挑选出来。
代码实现:
#include <iostream>
using namespace std;
const int N = 510, M = 510 * 110 / 2;
int n;
int arr[N];
int dp[N][M];
int main()
{cin >> n;int sum = 0;for (int i = 1; i <= n; i++){cin >> arr[i];sum += arr[i];}if (sum % 2 == 1) cout << "false" << endl;else{sum /= 2;dp[0][0] = true;for (int i = 1; i <= n; i++){for (int j = 0; j <= sum; j++){dp[i][j] = dp[i - 1][j];if (j >= arr[i]){dp[i][j] = dp[i][j] || dp[i - 1][j - arr[i]];}}}if (dp[n][sum]) cout << "true" << endl;else cout << "false" << endl;}return 0;
}