AtCoder Beginner Contest 172 總結

頹了三天端午回來打比賽，還是秒切前幾道水題但做不動後面的難題，任重道遠啊。

前四題思路都比較簡單，C枚舉一個數組的前綴和並用差值在另外一個數組裏二分查找，D可以線性篩約數個數函數然後暴力算，也可以分析每個數的貢獻來搞，複雜度都爲O(n)。

E沒有做出來是真的不爽，只想到了先考慮A，枚舉一個排列，A確定以後對B的計算方式確實沒有想到。看了題解說的容斥原理恍然大悟。先不考慮A算B有多少種排法（也是），然後減去至少有一位與A相同的排法（），再加回來至少兩位與A相同的排法（）......

當年搞OI時數學的薄弱板塊容斥果然沒有放過本人......

F有關博弈論，涉及的Nim遊戲還不太會，補了坑再回來寫。

 

 

A題

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

inline int read() {
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}

int main() {
	int n=read();
	printf("%d\n",n+n*n+n*n*n);
	return 0;
}

B題

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
string S,T;
inline int read() {
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}

int main() {
	cin>>S>>T;
	int n=S.length();
	int ans=0;
	for (register int i=0;i<n;++i)
		ans+=(S[i]!=T[i]);
	cout<<ans<<endl;
	return 0;
}

C題

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2e5+4;
int n,m;
ll K;
ll a[N],b[N];
inline int read() {
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
inline void smax(int &x,int y) {
	x=x<y?y:x;
}
int main() {
	n=read(),m=read(),K=read();
	for (register int i=1;i<=n;++i) a[i]=a[i-1]+read();
	for (register int i=1;i<=m;++i) b[i]=b[i-1]+read();
	int ans=0;
	for (register int i=0;i<=n;++i) {
		if (a[i]>K) break;
		int res=K-a[i];
		int pos=upper_bound(b+1,b+m+1,res)-b-1;
		smax(ans,i+pos);
	}
	printf("%d\n",ans);
	return 0;
}

D題

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e7+4;
typedef long long ll;
int n;
int prime[N],tot;
bool vis[N];
int d[N],t[N];
inline int read() {
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}

inline void linear_shaker(int lim) {
	d[1]=1;
	for (register int i=2;i<=lim;i++) {
		if (!vis[i]) prime[++tot]=i,d[i]=2,t[i]=1;
		for (register int j=1;j<=tot&&i*prime[j]<=lim;j++) {
			vis[i*prime[j]]=true;
			if (i%prime[j]==0) {
				t[i*prime[j]]=t[i]+1;
				d[i*prime[j]]=d[i]/(t[i]+1)*(t[prime[j]*i]+1);
				break;
			}
			t[i*prime[j]]=1;
			d[i*prime[j]]=d[i]*2;
		}
	}
}
ll ans;
int main() {
	n=read();
	linear_shaker(n);
	for (register int i=1;i<=n;++i)
		ans+=1ll*i*d[i];
	cout<<ans<<endl;
	return 0;
}

E題

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=5e5+5;
const ll MOD=1e9+7;
inline int read() {
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
ll fac[N]={1,1};
ll inv[N]={1,1};
int n,m;
inline ll fpow(ll a,int b) {
	ll ret=1;
	while (b) {
		if (b&1) ret=ret*a%MOD;
		b>>=1,a=a*a%MOD;
	}
	return ret;
}
inline ll C(int x,int y) {
	ll ans=fac[x];
	ans=ans*inv[y]%MOD;
	ans=ans*inv[x-y]%MOD;
	return ans;
}
inline ll P(int x,int y) {
	ll ans=fac[x];
	ans=ans*inv[x-y]%MOD;
	return ans;
}
inline void init() {
	for (register int i=1;i<N;++i) fac[i]=fac[i-1]*i%MOD;
	inv[N-1]=fpow(fac[N-1],MOD-2);
	for (int i=N-2;~i;--i) inv[i]=inv[i+1]*(i+1)%MOD;
}
int main() {
	n=read(),m=read();
	init();
	ll f=P(m,n);
	ll g=P(m,n),mk=1;
	for (register int i=1;i<=n;++i) {
		g-=mk*C(n,i)*P(m-i,n-i);
		g=(g%MOD+MOD)%MOD;
		mk=-mk;
	}
	ll ans=f*g%MOD;
	cout<<ans<<endl;
	return 0;
}

把D題的標準算法也貼一下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e7+4;
typedef long long ll;
int n;
inline int read() {
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
ll ans;
int main() {
	n=read();
	for (register int i=1;i<=n;++i) {
		int cnt=n/i;
		int s=i,t=cnt*i;
		ans+=1ll*(s+t)*cnt/2;
	}
	cout<<ans<<endl;
	return 0;
}