Codeforces Round #485 (Div. 2)

A. Infinity Gauntlet

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems:

• the Power Gem of purple color,
• the Time Gem of green color,
• the Space Gem of blue color,
• the Soul Gem of orange color,
• the Reality Gem of red color,
• the Mind Gem of yellow color.

Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.

Input

In the first line of input there is one integer $\mathrm{nn (0\le n\le 6) —\; the\; number\; of\; Gems\; in\; Infinity\; Gauntlet.}$

In next $\mathrm{nn lines\; there\; are\; colors\; of\; Gems\; you\; saw.\; Words\; used\; for\; colors\; are: purple, green, blue, orange, red, yellow.\; It\; is\; guaranteed\; that\; all\; the\; colors\; are distinct.\; All\; colors\; are\; given\; in\; lowercase\; English\; letters.}$

Output

In the first line output one integer $\mathrm{mm (0\le m\le 6) —\; the\; number\; of\; absent\; Gems.}$

Then in $\mathrm{mm lines\; print\; the\; names\; of\; absent\; Gems,\; each\; on\; its\; own\; line.\; Words\; used\; for\; names\; are: Power, Time, Space, Soul, Reality, Mind.\; Names\; can\; be\; printed\; in\; any\; order.\; Keep\; the\; first\; letter\; uppercase,\; others\; lowercase.}$

Examples

input

Copy

4
red
purple
yellow
orange

output

Copy

2
Space
Time

input

Copy

0

output

Copy

6
Time
Space
Power
Reality
Mind
Soul

Note

In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space.

In the second sample Thanos doesn't have any Gems, so he needs all six.

 题意：6种石头对应六中颜色，然后输入颜色，求没出现的颜色对应的石头

直接set处理下

#include<bits/stdc++.h>
#define ll long long
using namespace std;
map<string,string> mp;
int main()
{
    int n;
    set<string> s;
    string p;
    mp["purple"]="Power";
    mp["green"]="Time";
    mp["blue"]="Space";
    mp["orange"]="Soul";
    mp["red"]="Reality";
    mp["yellow"]="Mind";
    map<string,string>::iterator ip;
    for(ip=mp.begin();ip!=mp.end();ip++)
    {
        s.insert(ip->second);
    }
    scanf("%d",&n);
    int m=n;
    while(n--)
    {
        cin>>p;
        s.erase(mp[p]);
    }
    printf("%d
",6-m);
    set<string>::iterator it;
    for(it=s.begin(); it!=s.end(); it++)
    {
        cout<<*it<<endl;
    }
}

B. High School: Become Human

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.

It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.

One of the popular pranks on Vasya is to force him to compare ${\mathrm{x with yx.\; Other\; androids\; can\; do\; it\; in\; milliseconds\; while\; Vasya\text{'}s\; memory\; is\; too\; small\; to\; store\; such\; big\; numbers.}}^{}$

Please help Vasya! Write a fast program to compare ${\mathrm{xy with yx for\; Vasya,\; maybe\; then\; other\; androids\; will\; respect\; him.}}^{}$

Input

On the only line of input there are two integers $\mathrm{xx and yy (1\le x,y\le 109).}$

Output

If ${\mathrm{xy<yx,\; then\; print\; \text{'}<\text{'}\; (without\; quotes).\; If xy>yx,\; then\; print\; \text{'}>\text{'}\; (without\; quotes).\; If xy=yx,\; then\; print\; \text{'}=\text{'}\; (without\; quotes).}}^{}$

Examples

input

Copy

5 8

output

Copy

>

input

Copy

10 3

output

Copy

<

input

Copy

6 6

output

Copy

=

Note

In the first example ${\mathrm{58=5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=39062558=5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5=390625,\; and 85=8\cdot 8\cdot 8\cdot 8\cdot 8=3276885=8\cdot 8\cdot 8\cdot 8\cdot 8=32768.\; So\; you\; should\; print\; \text{'}>\text{'}.}}^{}$

In the second example ${\mathrm{103=1000<310=59049103=1000<310=59049.}}^{}$

In the third example ${\mathrm{66=46656=6666=46656=66.}}^{}$

题意：求x的y次和y的x次大小

直接判断log(x)*y与log(y)*x大小就行 我用python搞的emmm

import math
a,b=map(int,input().split())
aaa = b*math.log(a)
bbb = a*math.log(b)
if aaa>bbb:
    print ('>')
elif aaa<bbb:
    print ('<')
else :
    print ("=")

C. Three displays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.

There are $\mathrm{nn displays\; placed\; along\; a\; road,\; and\; the ii-th\; of\; them\; can\; display\; a\; text\; with\; font\; size sisi only.\; Maria\; Stepanovna\; wants\; to\; rent\; such\; three\; displays\; with\; indices i<j<ki<j<k that\; the\; font\; size\; increases\; if\; you\; move\; along\; the\; road\; in\; a\; particular\; direction.\; Namely,\; the\; condition si<sj<sksi<sj<sk should\; be\; held.}$

The rent cost is for the $\mathrm{ii-th\; display\; is cici.\; Please\; determine\; the\; smallest\; cost\; Maria\; Stepanovna\; should\; pay.}$

Input

The first line contains a single integer $\mathrm{nn (3\le n\le 3000) —\; the\; number\; of\; displays.}$

The second line contains $\mathrm{nn integers s1,s2,\dots ,sn(1\le si\le 109) —\; the\; font\; sizes\; on\; the\; displays\; in\; the\; order\; they\; stand\; along\; the\; road.}$

The third line contains $\mathrm{nn integers c1,c2,\dots ,cn (1\le ci\le 108) —\; the\; rent\; costs\; for\; each\; display.}$

Output

If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $\mathrm{i<j<ki<j<k such\; that si<sj<sksi<sj<sk.}$

Examples

input

Copy

5
2 4 5 4 10
40 30 20 10 40

output

Copy

90

input

Copy

3
100 101 100
2 4 5

output

Copy

-1

input

Copy

10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13

output

Copy

33

Note

In the first example you can, for example, choose displays $\mathrm{11, 44 and 55,\; because s1<s4<s5(2<4<102<4<10),\; and\; the\; rent\; cost\; is 40+10+40=9040+10+40=90.}$

In the second example you can't select a valid triple of indices, so the answer is -1.

 题意：输入n个数以及n个数的价值，然后求三个si递增的串，且ci和最小。

先确定第三位数dp[j]表示s[j]后面能取值最小的。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[5000],b[5000],dp[5000];
const int maxn=1e9+10;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%I64d",&a[i]);
    for(int i=1; i<=n; i++)
        scanf("%I64d",&b[i]);
    for(int i=1; i<=n; i++)
    {
        ll mn=maxn;
        for(int j=i+1; j<=n; j++)
        {
            if(a[i]<a[j])
            {
                mn=min(mn,b[j]);
            }
        }
        dp[i]=mn;
    }
    ll ans=maxn;
    for(int i=1; i<=n; i++)
    {
        for(int j=i+1; j<=n; j++)
        {
            if(a[i]<a[j])
            {
                if(dp[j]!=maxn)
                    ans=min(ans,b[i]+b[j]+dp[j]);
            }
        }
    }
    if(ans==maxn) printf("-1
");
    else cout<<ans<<endl;
}

D. Fair

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Some company is going to hold a fair in Byteland. There are $\mathrm{nn towns\; in\; Byteland\; and mm two-way\; roads\; between\; towns.\; Of\; course,\; you\; can\; reach\; any\; town\; from\; any\; other\; town\; using\; roads.}$

There are $\mathrm{kk types\; of\; goods\; produced\; in\; Byteland\; and\; every\; town\; produces\; only\; one\; type.\; To\; hold\; a\; fair\; you\; have\; to\; bring\; at\; least ssdifferent\; types\; of\; goods.\; It\; costs d(u,v)d(u,v) coins\; to\; bring\; goods\; from\; town uu to\; town vv where d(u,v)d(u,v) is\; the\; length\; of\; the\; shortest\; path\; from uuto vv.\; Length\; of\; a\; path\; is\; the\; number\; of\; roads\; in\; this\; path.}$

The organizers will cover all travel expenses but they can choose the towns to bring goods from. Now they want to calculate minimum expenses to hold a fair in each of $\mathrm{nn towns.}$

Input

There are $\mathrm{4 integers n, m, k, s in\; the\; first\; line\; of\; input\; (1\le n\le 105, 0\le m\le 105, 1\le s\le k\le min(n,100)) —\; the\; number\; of\; towns,\; the\; number\; of\; roads,\; the\; number\; of\; different\; types\; of\; goods,\; the\; number\; of\; different\; types\; of\; goods\; necessary\; to\; hold\; a\; fair.}$

In the next line there are $\mathrm{nn integers a1,a2,\dots ,an(1\le ai\le k),\; where aiai is\; the\; type\; of\; goods\; produced\; in\; the ii-th\; town.\; It\; is\; guaranteed\; that\; all\; integers\; between 11 and kk occur\; at\; least\; once\; among\; integers aiai.}$

In the next $\mathrm{mm lines\; roads\; are\; described.\; Each\; road\; is\; described\; by\; two\; integers uu vv (1\le u,v\le n u\ne v) —\; the\; towns\; connected\; by\; this\; road.\; It\; is\; guaranteed\; that\; there\; is\; no\; more\; than\; one\; road\; between\; every\; two\; towns.\; It\; is\; guaranteed\; that\; you\; can\; go\; from\; any\; town\; to\; any\; other\; town\; via\; roads.}$

Output

Print $\mathrm{nn numbers,\; the ii-th\; of\; them\; is\; the\; minimum\; number\; of\; coins\; you\; need\; to\; spend\; on\; travel\; expenses\; to\; hold\; a\; fair\; in\; town ii.\; Separate\; numbers\; with\; spaces.}$

Examples

input

Copy

5 5 4 3
1 2 4 3 2
1 2
2 3
3 4
4 1
4 5

output

Copy

2 2 2 2 3 

input

Copy

7 6 3 2
1 2 3 3 2 2 1
1 2
2 3
3 4
2 5
5 6
6 7

output

Copy

1 1 1 2 2 1 1 

Note

Let's look at the first sample.

To hold a fair in town $\mathrm{1 you\; can\; bring\; goods\; from\; towns 1 (0 coins), 2 (1 coin)\; and 4 (1 coin).\; Total\; numbers\; of\; coins\; is 2.}$

Town $\mathrm{2:\; Goods\; from\; towns 2 (0), 1 (1), 3 (1).\; Sum\; equals 2.}$

Town $\mathrm{3:\; Goods\; from\; towns 3 (0), 2 (1), 4 (1).\; Sum\; equals 2.}$

Town $\mathrm{4:\; Goods\; from\; towns 4 (0), 1 (1), 5 (1).\; Sum\; equals 2.}$

Town $\mathrm{5:\; Goods\; from\; towns\; 5 (0), 4 (1), 3 (2).\; Sum\; equals 3.}$

 题意：n个城镇,m条路(无向图),k种货物，每次举办活动需要s种货物。

按每种货物bfs，到每个城镇的最短路，然后排序，求最短的s条和即可。

#include<bits/stdc++.h>
#include<cmath>
#define ll long long
using namespace std;
const int maxn=100010;
const int INF = 0x3f3f3f3f;
int a[maxn],dis[110][maxn],n;
int head[maxn],To[maxn*200],Next[maxn*200],cnt,d[maxn];
void add(int u,int v)
{
    Next[++cnt]=head[u];
    head[u]=cnt;
    To[cnt]=v;
}
void bfs(int x)
{
    queue<int> q;
    for(int i=1; i<=n; i++)
        if(a[i]==x)
        {
            q.push(i);
            dis[x][i]=0;
        }
    while(!q.empty())
    {
        int v=q.front();
        q.pop();
        for(int i=head[v]; i; i=Next[i])
        {
            if(dis[x][To[i]]>dis[x][v]+1)
            {
                dis[x][To[i]]=dis[x][v]+1;
                q.push(To[i]);
            }
        }
    }

}
int main()
{
    int m,k,s,u,v;
    scanf("%d %d %d %d",&n,&m,&k,&s);
    for(int i=1; i<=n; i++) scanf("%d",&a[i]);
    memset(dis,INF,sizeof(dis));
    while(m--)
    {
        scanf("%d %d",&u,&v);
        add(u,v);
        add(v,u);
    }
    for(int i=1; i<=k; i++) bfs(i);
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=k; j++)
            d[j]=dis[j][i];
        sort(d+1,d+1+k);
        int res=0;
        for(int j=1; j<=s; j++)
            res+=d[j];
        printf("%d ",res);
    }

}

E. Petr and Permutations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petr likes to come up with problems about randomly generated data. This time problem is about random permutation. He decided to generate a random permutation this way: he takes identity permutation of numbers from $\mathrm{11 to nn and\; then 3n3n times\; takes\; a\; random\; pair\; of\; different\; elements\; and\; swaps\; them.\; Alex\; envies\; Petr\; and\; tries\; to\; imitate\; him\; in\; all\; kind\; of\; things.\; Alex\; has\; also\; come\; up\; with\; a\; problem\; about\; random\; permutation.}$

$\mathrm{He\; generates\; a\; random\; permutation\; just\; like\; Petr\; but\; swaps\; elements 7n+1times\; instead\; of 3n times.\; Because\; it\; is\; more\; random,\; OK?!}$

You somehow get a test from one of these problems and now you want to know from which one.

Input

In the first line of input there is one integer $\mathrm{n (103\le n\le 106).}$

In the second line there are $\mathrm{nn distinct\; integers\; between 1 and n —\; the\; permutation\; of\; size nn from\; the\; test.}$

It is guaranteed that all tests except for sample are generated this way: First we choose $\mathrm{nn —\; the\; size\; of\; the\; permutation.\; Then\; we\; randomly\; choose\; a\; method\; to\; generate\; a\; permutation —\; the\; one\; of\; Petr\; or\; the\; one\; of\; Alex.\; Then\; we\; generate\; a\; permutation\; using\; chosen\; method.}$

Output

If the test is generated via Petr's method print "Petr" (without quotes). If the test is generated via Alex's method print "Um_nik" (without quotes).

Example

input

Copy

5
2 4 5 1 3

output

Copy

Petr

Note

Please note that the sample is not a valid test (because of limitations for $\mathrm{n)\; and\; is\; given\; only\; to\; illustrate\; input/output\; format.\; Your\; program still\; has\; to\; print\; correct\; answer\; to\; this\; test to\; get\; AC.}$

Due to randomness of input hacks in this problem are forbidden.

 题意：给出一个长度为n的序列，原始序列应该是1 2 3 4 ...n ，但是被打乱了，Petr打乱的次数3n倍,Um_nik是7n+1.

因为7n+1与3n奇偶性不同，求出最小交换次数，判断就行。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
using namespace std;
const int maxn=1000010;
int n,pos[maxn],a[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),pos[a[i]]=i;
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]!=i)
        {
            int j=pos[i];
            swap(a[i],a[j]);
            pos[a[i]]=i;
            pos[a[j]]=j;
            cnt++;
        }
    }
    if(n%2==0)
    {
        if(cnt%2==0)
        printf("Petr");
        else printf("Um_nik");
    }
    else
    {
        if(cnt%2)
            printf("Petr");
        else printf("Um_nik");
    }

}

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~