题目1 : Visiting Peking University
时间限制:1000ms
单点时限:1000ms
内存限制:256MB
描述
Ming is going to travel for n days and the date of these days can be represented by n integers: 0, 1, 2, …, n-1. He plans to spend m consecutive days（2 ≤ m ≤ n）in Beijing. During these m days, he intends to use the first day and another day to visit Peking university. Before he made his plan, Ming investigated on the number of tourists who would be waiting in line to enter Peking university during his n-day trip, and the results could be represented by an integer sequence p[i] (0 ≤ i ≤ n-1, p[i] represents the number of waiting tourists on day i). To save time, he hopes to choose two certain dates a and b to visit PKU(0 ≤ a < b ≤ n-1), which makes p[a] + p[b] as small as possible.
Unfortunately, Ming comes to know that traffic control will be taking place in Beijing on some days during his n-day trip, and he won’t be able to visit any place in Beijing, including PKU, on a traffic control day. Ming loves Beijing and he wants to make sure that m days can be used to visit interesting places in Beijing. So Ming made a decision: spending k (m ≤ k ≤ n) consecutive days in Beijing is also acceptable if there are k - m traffic control days among those k days. Under this complicated situation, he doesn’t know how to make the best schedule. Please write a program to help Ming determine the best dates of the two days to visit Peking University. Data guarantees a unique solution.
输入
There are no more than 20 test cases.
For each test case:
The first line contains two integers, above mentioned n and m (2 ≤ n ≤ 100, 2 ≤ m ≤ n).
The second line contains n integers, above mentioned p[0] , p[1] , … p[n-1]. (0 ≤ p[i] ≤ 1000, i = 0 ... n-1)
The third line is an integer q (0 ≤ q ≤ n), representing the total number of traffic control days during the n-day trip, followed by q integers representing the dates of these days.
输出
One line, including two integers a and b, representing the best dates for visiting PKU.
- 样例输入
-
7 3
6 9 10 1 0 8 35
3 5 6 2
4 2
10 11 1 2
1 2
- 样例输出
-
0 3
1 3
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int a[101],i,j,n,m,x,y,s,h,u,v,k;
while(~scanf("%d%d",&n,&m))
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
scanf("%d",&x);
for(i=0;i<x;i++)
{
scanf("%d",&y);
a[y]=-1;
}
s=10000;u=v=0;x=y=0;
for(i=0;i<n;i++)
{
if(a[i]==-1)
continue;
k=m;
h=a[i];
for(j=i+1;j<k+i&&j<n;j++)
{
if(a[j]==-1)
{
k++;
continue;
}
if(h+a[j]<s)
{
s=h+a[j];
u=i,v=j;
}
}
if(k+i>n)
break;
else x=u,y=v;
}
printf("%d %d\n",x,y);
}
}
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=2e5+5;
int amax[maxn<<2];
int amin[maxn<<2];
int b[maxn];
const int inf=1<<30;
void pushup_max(int rt)
{
amax[rt]=max(amax[rt<<1],amax[rt<<1|1]);
}
void pushup_min(int rt)
{
amin[rt]=min(amin[rt<<1],amin[rt<<1|1]);
}
void build_max(int l,int r,int rt)
{
if(l==r)
{
amax[rt]=b[l];
return ;
}
int m=(l+r)>>1;
build_max(lson);
build_max(rson);
pushup_max(rt);
}
void build_min(int l,int r,int rt)
{
if(l==r)
{
amin[rt]=b[l];
return ;
}
int m=(l+r)>>1;
build_min(lson);
build_min(rson);
pushup_min(rt);
}
void update_max(int pos,int bue,int l,int r,int rt)
{
if(l==r)
{
amax[rt]=bue;
return;
}
int m=(l+r)>>1;
if(pos<=m) update_max(pos,bue,lson);
else update_max(pos,bue,rson);
pushup_max(rt);
}
void update_min(int pos,int bue,int l,int r,int rt)
{
if(l==r)
{
amin[rt]=bue;
return ;
}
int m=(l+r)>>1;
if(pos<=m) update_min(pos,bue,lson);
else update_min(pos,bue,rson);
pushup_min(rt);
}
int query_max(int left,int right,int l,int r,int rt)
{
if(left<=l&&right>=r)
{
return amax[rt];
}
int m=(l+r)>>1;
int t=-inf;
if(left<=m) t=max(t,query_max(left,right,lson));
if(right>m) t=max(t,query_max(left,right,rson));
return t;
}
int query_min(int left,int right,int l,int r,int rt)
{
if(left<=l&&right>=r)
{
return amin[rt];
}
int m=(l+r)>>1;
int t=inf;
if(left<=m) t=min(t,query_min(left,right,lson));
if(right>m) t=min(t,query_min(left,right,rson));
return t;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
n=1<<n;
for(int i=1; i<=n; i++)
scanf("%d",b+i);
build_max(1,n,1);
build_min(1,n,1);
int m;
scanf("%d",&m);
for(int i=0; i<m; i++)
{
int x,l,r;
scanf("%d%d%d",&x,&l,&r);
if(x==1)
{
int mi=query_min(++l,++r,1,n,1);
int ma=query_max(l,r,1,n,1);
if(ma*1LL*mi<=0)printf("%lld\n",ma*1LL*1LL*mi);
else
printf("%lld\n",min(ma*1LL*ma,mi*1LL*mi));
}
else
{
update_max(++l,r,1,n,1);
update_min(l,r,1,n,1);
}
}
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(int argc, char *argv[]) {
cin.sync_with_stdio(false);
ll n, m;
while (cin >> n >> m) {
if (n < m) swap(n, m);
ll gcd = __gcd(n - 1, m - 1);
ll k = (m - 1) / gcd;
ll x = (n - 1) / gcd;
ll ans = (k - 1) * (m - k + 1) + (x - k + 1) * (m - k);
cout << ans + 1 << endl;
}
return 0;
}
