A. Nearest Minimums
time limit per test
memory limit per test
input
output
You are given an array of n integer numbers a0, a1, ..., an - 1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.
Input
The first line contains positive integer n (2 ≤ n ≤ 105) — size of the given array. The second line contains n integers a0, a1, ..., an - 1 (1 ≤ ai ≤ 109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times.
Output
Print the only number — distance between two nearest minimums in the array.
Examples
Input
2 3 3
Output
1
Input
3 5 6 5
Output
2
Input
9 2 1 3 5 4 1 2 3 1
Output
3
思路：输入的时候搜一下最小值并记录一下位置，然后再找出最小位置
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[100005];
int b[100000];
int main()
{
int n;
cin>>n;
int mi=1000000000;
int ans=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]<mi){mi=a[i];ans=0;b[ans++]=i;}
else if(a[i]==mi){b[ans++]=i;}
}
int sum=1000000000;
for(int i=1;i<ans;i++)
{
if((b[i]-b[i-1])<sum)sum=b[i]-b[i-1];
}
cout<<sum<<endl;
return 0;
}
B. Two Cakes
time limit per test
memory limit per test
input
output
It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into a pieces, and the second one — into b
Ivan knows that there will be n people at the celebration (including himself), so Ivan has set n
- Each piece of each cake is put on some plate;
- Each plate contains at least one piece of cake;
- No plate contains pieces of both cakes.
To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number x such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least x
Help Ivan to calculate this number x!
Input
The first line contains three integers n, a and b (1 ≤ a, b ≤ 100, 2 ≤ n ≤ a + b) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.
Output
Print the maximum possible number x such that Ivan can distribute the cake in such a way that each plate will contain at least x
Examples
Input
5 2 3
Output
1
Input
4 7 10
Output
3
思路：按题目要求搜索一下
代码：
#include <iostream>
using namespace std;
int main()
{
int n, a, b;
cin >> n >> a >> b;
int sum1 = min(a, b);
int sum2 = max(a, b);
int ans = 1,mi = 0;
while (ans < n) {
int cx = n - ans;
int smc = sum1/ans;
int mxc = sum2/cx;
mi = max(mi,min(smc,mxc));
ans++;
}
cout << mi;
return 0;
}
C. Three Garlands
time limit per test
memory limit per test
input
output
Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if i-th garland is switched on during x-th second, then it is lit only during seconds x, x + ki, x + 2ki, x + 3ki
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers x1, x2 and x3 (not necessarily distinct) so that he will switch on the first garland during x1-th second, the second one — during x2-th second, and the third one — during x3-th second, respectively, and during each second starting from max(x1, x2, x3)
Help Mishka by telling him if it is possible to do this!
Input
The first line contains three integers k1, k2 and k3 (1 ≤ ki ≤ 1500) — time intervals of the garlands.
Output
If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO.
Examples
Input
2 2 3
Output
YES
Input
4 2 3
Output
NO
思路：找规律，一共四种情况
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[1550];
int main()
{
int q,w,e;
cin>>q>>w>>e;
a[q]++;a[w]++;a[e]++;
if(a[1]>0||a[2]>=2||a[3]>=3||(a[4]>=2&&a[2]==1))
cout<<"YES"<<endl;
else cout<<"NO"<<endl;
return 0;
}
D. Inversion Counting
time limit per test
memory limit per test
input
output
A permutation of size n is an array of size n such that each integer from 1 to n occurs exactly once in this array. An inversion in a permutation p is a pair of indices (i, j) such that i > j and ai < aj. For example, a permutation [4, 1, 3, 2] contains 4 inversions: (2, 1), (3, 1), (4, 1), (4, 3).
You are given a permutation a of size n and m queries to it. Each query is represented by two indices l and r denoting that you have to reverse the segment [l, r] of the permutation. For example, if a = [1, 2, 3, 4] and a query l = 2, r = 4 is applied, then the resulting permutation is [1, 4, 3, 2].
After each query you have to determine whether the number of inversions is odd or even.
Input
The first line contains one integer n (1 ≤ n ≤ 1500) — the size of the permutation.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — the elements of the permutation. These integers are pairwise distinct.
The third line contains one integer m (1 ≤ m ≤ 2·105) — the number of queries to process.
Then m lines follow, i-th line containing two integers li, ri (1 ≤ li ≤ ri ≤ n) denoting that i-th query is to reverse a segment [li, ri]
Output
Print m lines. i-th of them must be equal to odd if the number of inversions in the permutation after i-th query is odd, and even
Examples
Input
3 1 2 3 2 1 2 2 3
Output
odd even
Input
4 1 2 4 3 4 1 1 1 4 1 4 2 3
Output
odd odd odd even
思路：先算一下逆序数的个数，然后交换的数的顺序必改变。然后输出即可，刚开始用了树状数组，麻烦了
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[1550];
int c[1550];
int lowbit(int x)
{
return x&(-x);
}
void add(int x,int v)
{
while(x<=1550)
{
c[x]+=v;
x+=lowbit(x);
}
}
long long sum(int x)
{
long long sum=0;
while(x>0)
{
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
int main()
{
int n,m;
while(~scanf("%d",&n))
{
long long ans=0;
memset(c,0,sizeof(c));
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
add(a[i],1);
ans+=i-sum(a[i]-1);
}
// cout<<ans<<endl;
scanf("%d",&m);
int q,w,vv,v1;
while(m--)
{
scanf("%d%d",&q,&w);
q--;w--;
int su=w-q;
if(su%2==0)su=su/2;
else su=(su+1)/2;
ans+=su;
//cout<<ans<<endl;
if(ans%2==0)cout<<"even"<<endl;
else cout<<"odd"<<endl;
}
}
return 0;
}
