Codeforces Round#630 div2 A~C题解

                                                                                                           A. Exercising Walk

 

 

Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat!

Initially, Alice's cat is located in a cell (x,y)(x,y) of an infinite grid. According to Alice's theory, cat needs to move:

• exactly aa steps left: from (u,v)(u,v) to (u−1,v)(u−1,v);
• exactly bb steps right: from (u,v)(u,v) to (u+1,v)(u+1,v);
• exactly cc steps down: from (u,v)(u,v) to (u,v−1)(u,v−1);
• exactly dd steps up: from (u,v)(u,v) to (u,v+1)(u,v+1).

Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 11 step left, 33 steps right and 22 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x1,x2]×[y1,y2][x1,x2]×[y1,y2], i.e. for every cat's position (u,v)(u,v) of a walk x1≤u≤x2x1≤u≤x2 and y1≤v≤y2y1≤v≤y2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+da+b+c+d unit moves (aa to the left, bb to the right, cc to the down, dd to the up). Alice can do the moves in any order. Her current position (u,v)(u,v) should always satisfy the constraints: x1≤u≤x2x1≤u≤x2, y1≤v≤y2y1≤v≤y2. The staring point is (x,y)(x,y).

You are required to answer tt test cases independently.

 Input

The first line contains a single integer tt (1≤t≤1031≤t≤103) — the number of testcases.

The first line of each test case contains four integers aa, bb, cc, dd (0≤a,b,c,d≤1080≤a,b,c,d≤108, a+b+c+d≥1a+b+c+d≥1).

The second line of the test case contains six integers xx, yy, x1x1, y1y1, x2x2, y2y2 (−108≤x1≤x≤x2≤108−108≤x1≤x≤x2≤108, −108≤y1≤y≤y2≤108−108≤y1≤y≤y2≤108).

Output

For each test case, output "YES" in a separate line, if there exists a walk satisfying her wishes. Otherwise, output "NO" in a separate line.

You can print each letter in any case (upper or lower).

Example

input

6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0

output

Yes
No
No
Yes
Yes
Yes

题意：给你一个初始点(x,y),要求x1<=x<=x2,y1<=y<=y2,求这个点能否在这个活动范围内向左右上下移动a,b,c,d次
题解：因为左右移动和上下移动是独立的，所以我们将左右移动和上下移动分开来看,加入这个活动的范围不是空集,那么我们总可以向左(上)走一步然后向右(下)走一步,然后判断多出来的步数是否在区间范围内即可

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <stack>
 7 #include <queue>
 8 #include <vector>
 9 #include <map>
10 #include <set>
11 #include <unordered_set>
12 #include <unordered_map>
13 #define ll long long
14 #define fi first
15 #define se second
16 #define pb push_back
17 #define me memset
18 const int N = 1e6 + 10;
19 const int M = 1e9 + 7;
20 using namespace std;
21 typedef pair<int,int> PII;
22 typedef pair<long,long> PLL;
23
24 int t;
25 ll a,b,c,d;
26 ll x,y,x1,x2,y2;
27 ll Y1;
28 int main() {
29     ios::sync_with_stdio(false);
30        cin>>t;
31         while(t--) {
32             cin >> a >> b >> c >> d;
33             cin >> x >> y >> x1 >> Y1 >> x2 >> y2;
34             if (x1 == x2 && (a || b)) {
35                 printf("No\n");
36                 continue;
37             }
38             if (Y1 == y2 && (c || d)) {
39                 printf("No\n");
40                 continue;
41             }
42             x+=b-a;
43             y+=d-c;
44             if(x>=x1 && x<=x2 && y>=Y1 && y<=y2) printf("Yes\n");
45             else printf("No\n");
46         }
47     return 0;
48 }

                                                                     　　　　　　　    　B. Composite Coloring

A positive integer is called composite if it can be represented as a product of two positive integers, both greater than 11. For example, the following numbers are composite: 66, 44, 120120, 2727. The following numbers aren't: 11, 22, 33, 1717, 9797.

Alice is given a sequence of nn composite numbers a1,a2,…,ana1,a2,…,an.

She wants to choose an integer m≤11m≤11 and color each element one of mm colors from 11 to mm so that:

• for each color from 11 to mm there is at least one element of this color;
• each element is colored and colored exactly one color;
• the greatest common divisor of any two elements that are colored the same color is greater than 11, i.e. gcd(ai,aj)>1gcd(ai,aj)>1 for each pair i,ji,j if these elements are colored the same color.

Note that equal elements can be colored different colors — you just have to choose one of mm colors for each of the indices from 11 to nn.

Alice showed already that if all ai≤1000ai≤1000 then she can always solve the task by choosing some m≤11m≤11.

Help Alice to find the required coloring. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some mm from 11 to 1111.

Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of test cases. Then the descriptions of the test cases follow.

The first line of the test case contains a single integer nn (1≤n≤10001≤n≤1000) — the amount of numbers in a sequence aa.

The second line of the test case contains nn composite integers a1,a2,…,ana1,a2,…,an (4≤ai≤10004≤ai≤1000).

It is guaranteed that the sum of nn over all test cases doesn't exceed 104104.

Output

For each test case print 22 lines. The first line should contain a single integer mm (1≤m≤111≤m≤11) — the number of used colors. Consider colors to be numbered from 11 to mm. The second line should contain any coloring that satisfies the above conditions. Print nn integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤m1≤ci≤m), where cici is the color of the ii-th element. If there are multiple solutions then you can print any of them. Note that you don't have to minimize or maximize the number of colors, you just have to find the solution with some mm from 11 to 1111.

Remember that each color from 11 to mm should be used at least once. Any two elements of the same color should not be coprime (i.e. their GCD should be greater than 11).

Example

input

3
3
6 10 15
2
4 9
23
437 519 865 808 909 391 194 291 237 395 323 365 511 497 781 737 871 559 731 697 779 841 961

output

1
1 1 1
2
2 1
11
4 7 8 10 7 3 10 7 7 8 3 1 1 5 5 9 2 2 3 3 4 11 6

题意：给你n个合数,要求用m<=11种颜色来给他们图上颜色,每个数只能涂一种颜色,如果任意两个数的gcd>1，那么他们可以涂同一种颜色，相同的数也可以涂同一种颜色,要求输出颜色总数和种类
题解：该题的数据范围非常的小,因为每个数在1000以内,所以他们的质因数只有可能是(2，3，5，7，11，13，17，19，23，29，31),然后我们找每个数的最小质因数,如果他们的最小质因数相同，那么颜色也就相同,反之颜色种类++;

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <stack>
 7 #include <queue>
 8 #include <vector>
 9 #include <map>
10 #include <set>
11 #include <unordered_set>
12 #include <unordered_map>
13 #define ll long long
14 #define fi first
15 #define se second
16 #define pb push_back
17 #define me memset
18 const int N = 1e6 + 10;
19 const int M = 1e9 + 7;
20 using namespace std;
21 typedef pair<int,int> PII;
22 typedef pair<long,long> PLL;
23 int t;
24 int n,a[N],c[N];
25 int p[11]={2,3,5,7,11,13,17,19,23,29,31};
26 map<int,int> ans;
27
28 int main() {
29     ios::sync_with_stdio(false);
30         cin>>t;
31          while(t--){
32              me(c,0,sizeof(c));
33              ans.clear();
34              cin>>n;
35               for(int i=1;i<=n;++i) cin>>a[i];
36               int cnt=0;
37               for(int i=1;i<=n;++i)
38                   for(int j=0;j<11;++j){
39                       if(a[i]%p[j]==0){
40                           if(c[j]==0) c[j]=++cnt;
41                           ans[i]=c[j];
42                           break;
43                       }
44                   }
45               map<int,int>::iterator iter;
46                   printf("%d\n",cnt);
47                   for(iter=ans.begin();iter!=ans.end();++iter) printf("%d ",iter->se);
48                   printf("\n");
49          }
50     return 0;
51 }

 

 　　　　　　　　　　　　　　　　　　　　　　　　　　　 　C. K-Complete Word

Word ss of length nn is called kk-complete if

• ss is a palindrome, i.e. si=sn+1−isi=sn+1−i for all 1≤i≤n1≤i≤n;
• ss has a period of kk, i.e. si=sk+isi=sk+i for all 1≤i≤n−k1≤i≤n−k.

For example, "abaaba" is a 33-complete word, while "abccba" is not.

Bob is given a word ss of length nn consisting of only lowercase Latin letters and an integer kk, such that nn is divisible by kk. He wants to convert ss to any kk-complete word.

To do this Bob can choose some ii (1≤i≤n1≤i≤n) and replace the letter at position ii with some other lowercase Latin letter.

So now Bob wants to know the minimum number of letters he has to replace to convert ss to any kk-complete word.

Note that Bob can do zero changes if the word ss is already kk-complete.

You are required to answer tt test cases independently.

Input

The first line contains a single integer tt (1≤t≤1051≤t≤105) — the number of test cases.

The first line of each test case contains two integers nn and kk (1≤k<n≤2⋅1051≤k<n≤2⋅105, nn is divisible by kk).

The second line of each test case contains a word ss of length nn.

It is guaranteed that word ss only contains lowercase Latin letters. And it is guaranteed that the sum of nn over all test cases will not exceed 2⋅1052⋅105.

Output

For each test case, output one integer, representing the minimum number of characters he has to replace to convert ss to any kk-complete word.

Example

input

4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp

output

2
0
23
16

题意：给你一个长度为n的字符串s，要求对于1<=i+k<=n,有s[i]=s[i+k],且s是一个回文串,问要达到这样的要求,至少要修改多少个字母(n可以被k整除)

  题解：因为s必须是回文串,所以,要把s改成n/k个相等的串,并且每个串必须是回文串,我们可以直接遍历得出长度为k的串的前k/2的位置上的每一个字符在s中出现的次数,然后再找与之相对应的回文位置的字符在s中出现的次数,取max,2*n/k-max(*2是因为还要确保回文,所以相               对回文的位置也要修改)即是修改的次数,

            注意:如果k是奇数,不要忘了最中间的那个字符,所以此时我们还要对中间位置k/2再找一次max,用原来的次数加上n/k-max(因为这次只找了一个位置,所以不用*2)就是答案了

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <stack>
 7 #include <queue>
 8 #include <vector>
 9 #include <map>
10 #include <set>
11 #include <unordered_set>
12 #include <unordered_map>
13 #define ll long long
14 #define fi first
15 #define se second
16 #define pb push_back
17 #define me memset
18 const int N = 1e6 + 10;
19 const int M = 1e9 + 7;
20 using namespace std;
21 typedef pair<int,int> PII;
22 typedef pair<long,long> PLL;
23
24 int t;
25 int n,k;
26 string s;
27 int cnt[26];
28 int main() {
29     ios::sync_with_stdio(false);
30     cin>>t;
31      while(t--){
32          cin>>n>>k>>s;
33          int ans=0;
34            for(int i=0;i<k/2;++i){
35                int ma=0;
36                me(cnt,0,sizeof(cnt));
37                for(int j=0;j<n/k;++j){
38                    cnt[s[j*k+i]-'a']++;
39                    ma=max(ma,cnt[s[j*k+i]-'a']);
40                    cnt[s[j*k+(k-i-1)]-'a']++;
41                    ma=max(ma,cnt[s[j*k+(k-i-1)]-'a']);
42                }
43                ans+=2*(n/k)-ma;
44            }
45
46            if(k%2==1){
47                me(cnt,0,sizeof(cnt));
48                int ma=0;
49                int i=k/2;
50                for(int j=0;j<n/k;++j){
51                    cnt[s[j*k+i]-'a']++;
52                    ma=max(ma,cnt[s[j*k+i]-'a']);
53                }
54                ans+=n/k-ma;
55            }
56            printf("%d\n",ans);
57      }
58
59
60
61     return 0;
62 }