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\n",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\n",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\n";
//printmat(mat);
//cout<<"^"<<p<<"\n";
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:\n";
//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\n",a);
continue;
}
if(n==2){printf("%lld\n",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<<"\n";
//cout<<"MAT=\n";
//printmat(mt);
l=r+1;
}
printf("%lld\n",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\n");
}
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\n",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\n",a,b);
}
return 0;
}
