Codeforces Round #452 (Div. 2) A-C题解

来源：互联网发布：sql创建视图编辑：程序博客网时间：2024/06/05 00:30

哎............

There were n groups of students which came to write a training contest. A group is either one person who can write the contest with anyone else, or two people who want to write the contest in the same team.

The coach decided to form teams of exactly three people for this training. Determine the maximum number of teams of three people he can form. It is possible that he can't use all groups to form teams. For groups of two, either both students should write the contest, or both should not. If two students from a group of two will write the contest, they should be in the same team.

Input

The first line contains single integer n (2 ≤ n ≤ 2·105) — the number of groups.

The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 2), where ai is the number of people in group i.

Output

Print the maximum number of teams of three people the coach can form.

Examples

input

41 1 2 1

output

input

22 2

output

input

72 2 2 1 1 1 1

output

input

31 1 1

output

Note

In the first example the coach can form one team. For example, he can take students from the first, second and fourth groups.

In the second example he can't make a single team.

In the third example the coach can form three teams. For example, he can do this in the following way:

The first group (of two people) and the seventh group (of one person),
The second group (of two people) and the sixth group (of one person),
The third group (of two people) and the fourth group (of one person).

组队，每队只有1或两个人，两个人的队要么全选，要么全不选，问最多能组成多少队。

代码实现：

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#include<set>#include<map>#include<cstdio>#define ll long long#define mset(a,x) memset(a,x,sizeof(a))using namespace std;const double PI=acos(-1);const int inf=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int mod=1e9+7;int dir[4][2]={0,1,1,0,0,-1,-1,0};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}int s[maxn];int vis[10];int main(){int i,j,k,n;while(~scanf("%d",&n)){mset(vis,0);int ans=0;for(i=0;i<n;i++){scanf("%d",&s[i]);vis[s[i]]++;}if(vis[1]<vis[2]){ans=vis[1];}else{ans=vis[2]+(vis[1]-vis[2])/3;}printf("%d\n",ans);}return 0;}

Everybody in Russia uses Gregorian calendar. In this calendar there are 31 days in January, 28 or 29 days in February (depending on whether the year is leap or not), 31 days in March, 30 days in April, 31 days in May, 30 in June, 31 in July, 31 in August, 30 in September, 31 in October, 30 in November, 31 in December.

A year is leap in one of two cases: either its number is divisible by 4, but not divisible by 100, or is divisible by 400. For example, the following years are leap: 2000, 2004, but years 1900 and 2018 are not leap.

In this problem you are given n (1 ≤ n ≤ 24) integers a1, a2, ..., an, and you have to check if these integers could be durations in days of n consecutive months, according to Gregorian calendar. Note that these months could belong to several consecutive years. In other words, check if there is a month in some year, such that its duration is a1 days, duration of the next month is a2 days, and so on.

Input

The first line contains single integer n (1 ≤ n ≤ 24) — the number of integers.

The second line contains n integers a1, a2, ..., an (28 ≤ ai ≤ 31) — the numbers you are to check.

Output

If there are several consecutive months that fit the sequence, print "YES" (without quotes). Otherwise, print "NO" (without quotes).

You can print each letter in arbitrary case (small or large).

Examples
input
431 31 30 31
output
Yes
input
230 30
output
No
input
529 31 30 31 30
output
Yes
input
331 28 30
output
No
input
331 31 28
output
Yes

Note

In the first example the integers can denote months July, August, September and October.

In the second example the answer is no, because there are no two consecutive months each having 30 days.

In the third example the months are: February (leap year) — March — April – May — June.

In the fourth example the number of days in the second month is 28, so this is February. March follows February and has 31 days, but not 30, so the answer is NO.

In the fifth example the months are: December — January — February (non-leap year).

给定连续的N个月，问存不存在这样的连续月份。

代码实现：

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#include<set>#include<map>#include<cstdio>#define ll long long#define mset(a,x) memset(a,x,sizeof(a))using namespace std;const double PI=acos(-1);const int inf=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int mod=1e9+7;int dir[4][2]={0,1,1,0,0,-1,-1,0};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}int s[100];int month1[]={31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,};int s1[100],s2[100];int main(){int n,i,j,k,flag=0;while(~scanf("%d",&n)){flag=0;for(i=0;i<n;i++){scanf("%d",&s[i]);}for(i=0;i<48;i++){int x=1,temp=i,y;for(j=0;j<n;j++){k=j%48;y=temp%48;if(month1[y]!=s[k]){x=0;break;}temp++;}if(x==1){cout<<"YES"<<endl;return 0;}}cout<<"NO"<<endl;}return 0;}

Petya has n integers: 1, 2, 3, ..., n. He wants to split these integers in two non-empty groups in such a way that the absolute difference of sums of integers in each group is as small as possible.

Help Petya to split the integers. Each of n integers should be exactly in one group.

Input

The first line contains a single integer n (2 ≤ n ≤ 60 000) — the number of integers Petya has.

Output

Print the smallest possible absolute difference in the first line.

In the second line print the size of the first group, followed by the integers in that group. You can print these integers in arbitrary order. If there are multiple answers, print any of them.

Examples

input

output

02 1 4

input

output

11 1

Note

In the first example you have to put integers 1 and 4 in the first group, and 2 and 3 in the second. This way the sum in each group is 5, and the absolute difference is 0.

In the second example there are only two integers, and since both groups should be non-empty, you have to put one integer in the first group and one in the second. This way the absolute difference of sums of integers in each group is 1.

1到N，分成两组，使得差的绝对值最小。输出绝对值，以及第一组数的数量和数据。

偶数的时候4的倍数都是0，其它规律可以多写几组就可以发现。

代码实现：

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#include<set>#include<map>#include<cstdio>#define ll long long#define mset(a,x) memset(a,x,sizeof(a))using namespace std;const double PI=acos(-1);const int inf=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int mod=1e9+7;int dir[4][2]={0,1,1,0,0,-1,-1,0};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}ll sum[100000];ll dp[60005];int s[60005];int main(){int n,i,j,k;sum[0]=0;for(i=1;i<=60000;i++){sum[i]+=sum[i-1]+i;s[i]=i;}scanf("%d",&n);if(n==2){printf("1\n1 1\n");return 0;}if(n%2==0){if(n%4==0){printf("0\n%d ",n/2);for(j=1;j<=n/2;j+=2){printf("%d %d ",j,n-j+1);}return 0;}printf("1\n%d ",n/2);for(j=1;j<n/2;j+=2){printf("%d %d ",j,n-j+1);}printf("%d",n/2);return 0;}else{ll a=0;for(i=1;i<=n;i+=4){if(i==n)a+=n;elsea+=(i+i+1);}printf("%lld\n",abs(sum[n]-2*a));printf("%d ",n/2+1);for(i=1;i<=n;i+=4){if(i==n)printf("%d ",i);elseprintf("%d %d ",i,i+1);}}return 0;}

阅读全文

0 0