Contest1143 - 計算思維實訓-ACM程序設計練習

點擊打開鏈接

#include<bits/stdc++.h>
using namespace std;
long long dp[10000];
void dpp(int n){
	dp[0] = 0;
	dp[1] = 1;
	for(int i = 2;i<=47;i++){
		dp[i] = dp[i-1]+dp[i-2];
	}
}
int main(){
	int n;
	dpp(47);
	while(cin>>n&&n!=-1){
		int yn  =0;
		for(int i = 0;i<=47;i++){
			if(dp[i] == n){
				cout<<i<<endl;
				yn = 1;
				break;
			}
		}
		if(yn == 0)cout<<"Not a Fibonacci number."<<endl;
	}
}

B: 點擊打開鏈接

#include<bits/stdc++.h>
using namespace std;
long long dp[66];
void dpp (){
	dp[1] =  3;
	dp[2] = 6;
	dp[3] = 6;
	for(int i = 4;i<62;i++){
		dp[i] = dp[i-1]+2*dp[i-2];
	}
}
int main(){
	int n;
	dpp();
	while(cin>>n){
		cout<<dp[n]<<endl;
	}
}

D:

   http://202.121.199.212/JudgeOnline/problem.php?cid=1143&pid=3

c數組輸出可以得到過河過程

#include<bits/stdc++.h>
using namespace std;
int a[1005];
int b[5];
int c[1000][2];
int main(){
    int n;
    while(cin>>n){
        for(int i = 0;i<n;i++)cin>>a[i];
        sort(a,a+n);
        if(n == 1){cout<<a[0]<<endl;continue;}
        //if(n == 2){cout<<a[1]<<endl;continue;}
        int t = n;
        b[0] = a[0];
        b[1] = a[1];
        int m = 1;
        long long sum = 0;
        int cnt = 0;
        while(t>0){
        if(n>2&&t>2){
            if(m == 1){
 
            //  cout<<b[1]<<" "<<b[0]<<endl;
            //  cout<<b[0]<<endl;
                c[++cnt][0]= b[0];
                c[cnt][1] = b[1];
                c[++cnt][0] = b[0];
                sum += b[1]+b[0];
                t=t-1;
                m = 0;
            }
            else{
 
 
                c[++cnt][0]= a[n-2];
                c[cnt][1] = a[n-1];
                c[++cnt][0] = b[1];
            //  cout<<a[n-2]<<" "<<a[n-1]<<endl;
            //  cout<<b[1]<<endl;
                sum+= a[n-1]+b[1];
                n-=2;
                t-=1;
                m++;
            }
        }
        else{
        if(m == 1){
    //  cout<<b[1]<<" "<<b[0]<<endl;
                sum += b[1];
                m = 0;
                t-=2;
                c[++cnt][0]= b[0];
                c[cnt][1] = b[1];
            }
            else {
                sum+= a[n-1];
            //  cout<<a[0]<<" "<<a[n-1]<<endl;
 
                c[++cnt][0]= b[0];
                c[cnt][1] = a[n-1];
                n-=2;
                t-=2;
                m++;
            }
        }
        //cout<<sum<<endl;
        }
        cout<<sum<<endl;
        /*for(int i = 1;i<=cnt;i++){
        if(c[i][1] != 0)
            cout<<c[i][0]<<" "<<c[i][1]<<endl;
            else  cout<<c[i][0]<<endl;
        }*/
    }
}

F:

點擊打開鏈接

  #include <iostream>
    #include<cstring>
    #include <algorithm>
    #define INF 0x3f3f3f3f
    using namespace std;
    const int N = 1e5 + 5;
    int s[N];
    int n,p,a[N];
    int len;
    int main()
    {
      //  cin>>n;
        while(cin>>p){
 
            memset(s,0,sizeof(s));
            for(int i = 0;i<p;i++)cin>>a[i];
            s[1] = a[0];len = 1;//長度從1開始
            for(int i = 1;i<p;i++){
 
                int t = a[i];
                if(t>s[len])s[++len] = a[i];
                else{
            /*************/int l = 1,r = len,mid;//這裏的二分法採用了左閉右閉的思路
int ans = 0;
                    while(l<=r)
                    {
                        mid = (l+r)/2;
                        if(s[mid]<t)
                            {l = mid +1;ans = max(ans,mid);}//ans即爲思路中的j，j必然爲s數組中小於t的最大的數
                        else r = mid-1;
                    }
                    s[ans+1] = t;/******************/
                }
            }
            //for(int i = 1;i<p;i++){cout<<s[i];}//有必要可以打開看看s中存的是什麼值
            cout<<len<<endl;
        }
        return 0;
    }