ZROI 菜鸡互啄杯 III

2018.8.14

讲题解以前，先膜拜一下 Siyuan 小姐姐家族的人
题面 0814.pdf

A.

数学水题
by sshwy & xry

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,t,mx;

int main(){scanf("%lld",&t);
	for(ll i=1;i<=t;i++){scanf("%lld",&n);
		mx=-1;
		ll x,y,z;
		//a==2
		if(n%2==0){
			x=n/2;
			ll n2=n-x;
			if(n%3==0){//b==3
				y=n/3,z=n-x-y;
				if(n%z==0)mx=max(mx,x*y*z);
			}
			if(n%4==0){//b==4
				y=n/4,z=n-x-y;
				if(n%z==0)mx=max(mx,x*y*z);
			}
		}		
		//a==3
		if(n%3==0){x=n/3,mx=max(mx,x*x*x);
		}
		printf("%lld
",mx);
	}
	return 0;
}

B.

乘法原理
by GavinZHeng & sshwy

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll t,n,m,b[200];
int main(){scanf("%lld",&t);
	while(t--){scanf("%lld%lld",&n,&m);
		for(ll i=1;i<=n;i++)scanf("%*d%lld",&b[i]),b[i]++;
		sort(b+1,b+n+1);
		ll ans=1,i;
		for(i=1;i<=n;i++){if(ans*b[i]>m)break;
			ans*=b[i];
			
		}
		printf("%lld
",i-1);
	}
	return 0;
}

C.

矩阵加速（加分块？）
by sry

#include<bits/stdc++.h>
using namespace std;
const long long mod=1e9+7;
struct matrix{long long r,c,m[5][5];
	matrix operator *(matrix t)const{
		matrix tmp;
		tmp.r=r,tmp.c=t.c;
		for(int i=1;i<=tmp.r;i++){for(int j=1;j<=tmp.c;j++){tmp.m[i][j]=0;
			}
		}
		for(int i=1;i<=r;i++){for(int j=1;j<=t.c;j++){for(int k=1;k<=c;k++){tmp.m[i][j]+=m[i][k]*t.m[k][j];
					tmp.m[i][j]%=mod;
				}
			}
		}
		return tmp;
	}
};
matrix mt,z;
void printmat(matrix x){for(int i=1;i<=x.r;i++){for(int j=1;j<=x.c;j++)cout<<x.m[i][j]<<" ";
		puts("");
	}
}
matrix POW(matrix mat,long long p){
	//cout<<"POW
";
	//printmat(mat);
	//cout<<"^"<<p<<"
";
	matrix tmp;
	tmp.r=mat.r,tmp.c=mat.c;
	for(int i=1;i<=tmp.r;i++){for(int j=1;j<=tmp.c;j++){if(i==j)tmp.m[i][j]=1;
			else tmp.m[i][j]=0;
		}
	}
	while(p){if(p%2)tmp=tmp*mat;
		mat=mat*mat,p/=2;
	}
	//cout<<"TMP:
";
	//printmat(tmp);
	return tmp;
}
long long T,a,b,c,d,p,n,l,r,v;
int main(){scanf("%lld",&T);
	while(T--){scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&d,&p,&n);
		if(n==1){printf("%lld
",a);
			continue;
		}
		if(n==2){printf("%lld
",b);
			continue;
		}
		mt.r=1;
		mt.c=3;
		mt.m[1][1]=1,mt.m[1][2]=a,mt.m[1][3]=b;
		z.c=z.r=3;
		z.m[1][1]=1,z.m[1][2]=0;
		z.m[2][1]=0,z.m[2][2]=0,z.m[2][3]=c;
		z.m[3][1]=0,z.m[3][2]=1,z.m[3][3]=d;
		l=3;
		while(l<=n){
			v=p/l;
			if(v)r=min(n,p/v);
			else r=n;
			z.m[1][3]=v,mt=mt*POW(z,r-l+1);
			//cout<<"l="<<l<<"r="<<r<<"
";
			//cout<<"MAT=
";
			//printmat(mt);
			l=r+1;
		}
		printf("%lld
",mt.m[1][3]);
	}
	return 0;
}

H.

不读题 + 瞎想系列
by sshwy & GavinZheng & zry

#include<bits/stdc++.h>
using namespace std;
int main(){
	int n;
	while(~scanf("%d",&n)){printf("Yes
");
	}
	return 0;
}

I.

逆序对
by sshwy

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=100005;
ll n,x,y,tot;
ll a[N],tmp[N];

void merge_sort(ll *arr,ll l,ll r){if(l==r)return;
	ll mid=(l+r)>>1;
	merge_sort(arr,l,mid);
	merge_sort(arr,mid+1,r);
	for(ll i=l,j=mid+1,k=l;i<=mid||j<=r;k++){if(i>mid)tmp[k]=arr[j],j++;
		else if(j>r||arr[i]<=arr[j])tmp[k]=arr[i],i++,tot+=j-mid-1;
		else tmp[k]=arr[j],j++;
	}
	for(ll i=l;i<=r;i++)arr[i]=tmp[i];
}

int main(){while(~scanf("%lld%lld%lld",&n,&x,&y)){for(ll i=1;i<=n;i++)scanf("%lld",&a[i]);
		tot=0;
		merge_sort(a,1,n);
		printf("%lld
",min(x,y)*tot);
	}
	return 0;
}//sshwy

J.

数据读入处理
by sshwy & GavinZheng

#include<bits/stdc++.h>
using namespace std;
int t,a,b;
char s[100];

void rd(int & x,int & y){
	x=y=0;
	char c=getchar();bool isN=0;
	while(!isdigit(c)){c=getchar();
		if(c=='-')isN=1;
	}
	while(isdigit(c))x=x*10+c-'0',c=getchar();
	if(isN)x=0-x;
	if(c!='.')return;
	c=getchar();
	while(isdigit(c))y=y*10+c-'0',c=getchar();
	if(isN)y=0-y;
}

int main(){scanf("%d",&t);
	while(t--){scanf("%d%d",&a,&b);
		int p,q;
		rd(p,q);
		q=q*6,b=(b+q);
		a=(a+p-8+(int)floor(b*1.0/60)+24)%24,b=(b+60)%60;
		printf("%02d:%02d
",a,b);
	}
	return 0;
}