AtCoder beginer const 178 A--F题解

A Not https://vjudge.net/problem/AtCoder-abc178_a
签到

#include<iostream>
#include<string>
#include<queue>
#include<algorithm>
#include<cstdlib>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int maxn=1e5+5;
int n,k;
ll a[maxn];
ll mp[maxn];
 
int main()
{
    int n;
    cin>>n;
    if(n==1) cout<<0<<endl;
    else cout<<1<<endl;
    return 0;
}

B Product Max https://vjudge.net/problem/AtCoder-abc178_b
签到

#include<iostream>
#include<string>
#include<queue>
#include<algorithm>
#include<cstdlib>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int maxn=1e5+5;
int n,k;
int main()
{
    ll a,b,c,d;
    cin>>a>>b>>c>>d;
    ll maxx=0;
    ll a1=a*c;
    ll b1=a*d;
    ll c1=b*c;
    ll d1=b*d;
    cout<<max(max(a1,b1),max(c1,d1))<<endl;
	system("pause");
    return 0;
}

C Ubiquity https://vjudge.net/problem/AtCoder-abc178_c
快速幂

#include<iostream>
#include<string>
#include<queue>
#include<algorithm>
#include<cstdlib>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int maxn=1e5+5;
const int mod=1e9+7;
 
inline ll quick(ll a,ll b,ll p)	//(a^b) mod p 
{
	ll ans=1;
	for( ;b;b>>=1)
	{
		if(b&1) ans=ans*a%p;
		a=a*a%p;
	}
	return ans%p;	//如果b=0,p=1,不加%p会错 
}
 
int main()
{
    ll n;
    cin>>n;
    ll t=(quick(10,n,mod)+quick(8,n,mod)-2*quick(9,n,mod)+2*mod)%mod;
    cout<<t<<endl;
	system("pause");
    return 0;
}

D Redistribution https://vjudge.net/problem/AtCoder-abc178_d
一开始以为是dp题，后来找到规律，有人一行代码过是真夸张D_D

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<stdlib.h>
 
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cstdlib>
 
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=1e6+6;
const int mod=1e9+7;
ll a[maxn]={1,0,0,1,1};
bool cmp(ll a,ll b){
    return a>b;
}
int main()
{
    cin.tie(0);std::ios::sync_with_stdio(false);
    int n;
    cin>>n;
    if(n>4){
        for(int i=5;i<=n;i++)
            a[i]=(a[i-1]+a[i-3])%mod;
    }
    cout<<a[n]<<endl;
	system("pause");
    return 0;
}

E Dist Max https://vjudge.net/problem/AtCoder-abc178_e
平面的曼哈顿距离，将绝对值x1-y1+绝对值x2-y2分类讨论，转化成max(a[n]-a[1],b[n]-b[1])

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<stdlib.h>
 
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cstdlib>
 
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=1e6+6;
const int mod=1e9+7;
ll a[maxn],b[maxn];
 
int main()
{
    cin.tie(0);std::ios::sync_with_stdio(false);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        int x,y;
        cin>>x>>y;
        a[i]=x+y;
        b[i]=x-y;
    }
    sort(a+1,a+1+n);
    sort(b+1,b+1+n);
    cout<<max(a[n]-a[1],b[n]-b[1])<<endl;
	system("pause");
    return 0;
}

F Contrast https://vjudge.net/problem/AtCoder-abc178_f
当时没做出来，看了份别人的代码，好像是把一个数组从前向后排，一个从后向前排，逐个进行比较如果有一样的就向后找一个不相同的swap过去
贴一份atcoder上的代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<stdlib.h>

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cstdlib>

#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=2e5+100;
const int mod=1e9+7;
int n,a[maxn],b[maxn];
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a[i];
	for(int i=n;i>=1;i--)
		cin>>b[i];
	for(int i=1;i<=n;i++){
		if(a[i]==b[i]){
			bool flag=0;
			for(int j=1;j<=n;j++){
				if(b[j]!=a[i]&&b[i]!=a[j]){
					swap(b[i],b[j]);
					flag=1;
					break;
				}
			}
			if(!flag){
				cout<<"No";
				return 0;
			}
		}
	}
	cout<<"Yes"<<endl;
	for(int i=1;i<=n;i++) cout<<b[i]<<" ";
	system("pause");
    return 0;
}