codeforces練習 - JavaShuo

Descriptionreact

DZY loves colors, and he enjoys painting.ios

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.app

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.ide

DZY wants to perform m operations, each operation can be one of the following:ui

Paint all the units with numbers between l and r (both inclusive) with color x.
Ask the sum of colorfulness of the units between l and r (both inclusive).

Can you help DZY?this

Inputspa

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 10⁵)..net

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.3d

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 10⁸) in this line, describing an operation 1.code

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Sample Input

Input

Output

Input

Output

3
2
1

Input

Output

Hint

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.

線段樹成段更新

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std ;
 5 
 6 #define ls ( o << 1 )
 7 #define rs ( o << 1 | 1 )
 8 #define lson ls , l , m
 9 #define rson rs , m + 1 , r
10 #define rt o , l , r
11 #define root 1 , 1 , n
12 #define mid ( ( l + r ) >> 1 )
13 #define clear( a , x ) memset ( a , x , sizeof a )
14 
15 typedef long long LL ;
16 
17 const int MAXN = 100005 ;
18 
19 int set[MAXN << 2] ;
20 LL sum[MAXN << 2] ;
21 LL add[MAXN << 2] ;
22 
23 void pushUp ( int o , int l , int r ) {
24     set[o] = ( set[ls] == set[rs] ? set[ls] : 0 ) ;
25     sum[o] = sum[ls] + sum[rs] ;
26 }
27 
28 void pushDown ( int o , int l , int r ) {
29     int m = mid ;
30     if ( set[o] ) set[ls] = set[rs] = set[o] ;
31     if ( add[o] ) {
32         sum[ls] += add[o] * ( m - l + 1 ) ;
33         add[ls] += add[o] ;
34         sum[rs] += add[o] * ( r - m ) ;
35         add[rs] += add[o] ;
36         add[o] = 0 ;
37     }
38 }
39 
40 void build ( int o , int l , int r ) {
41     add[o] = set[o] = sum[o] = 0 ;
42     if ( l == r ) {
43         set[o] = l ;
44         return ;
45     }
46     int m = mid ;
47     build ( lson ) ;
48     build ( rson ) ;
49 }
50 
51 void update ( int x , int L , int R , int o , int l , int r ) {
52     if ( L <= l && r <= R ) {
53         if ( set[o] ) {
54             add[o] += abs ( x - set[o] ) ;
55             sum[o] += ( LL ) abs ( x - set[o] ) * ( r - l + 1 ) ;
56             set[o] = x ;
57             return ;
58         }
59     }
60     pushDown ( rt ) ;
61     int m = mid ;
62     if ( L <= m ) update ( x , L , R , lson ) ;
63     if ( m <  R ) update ( x , L , R , rson ) ;
64     pushUp ( rt ) ;
65 }
66 
67 LL query ( int L , int R , int o , int l , int r ) {
68     if ( L <= l && r <= R ) return sum[o] ;
69     pushDown ( rt ) ;
70     int m = mid ;
71     LL ans = 0 ;
72     if ( L <= m ) ans += query ( L , R , lson ) ;
73     if ( m <  R ) ans += query ( L , R , rson ) ;
74     return ans ;
75 }
76 
77 void work () {
78     int n , m ;
79     int type , l , r , x ;
80     while ( ~scanf ( "%d%d" , &n , &m ) ) {
81         build ( root ) ;
82         while ( m -- ) {
83             scanf ( "%d%d%d" , &type , &l , &r ) ;
84             if ( type == 1 ) {
85                 scanf ( "%d" , &x ) ;
86                 update ( x , l , r , root ) ;
87             }
88             else printf ( "%I64d\n" , query ( l , r , root ) ) ;
89         }
90     }
91 }
92 
93 int main () {
94     work () ;
95     return 0 ;
96 }

View Code

DZY Loves Chessboard

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Appoint description:

Description

DZY loves chessboard, and he enjoys playing with it.

He has a chessboard of n rows and m columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.

You task is to find any suitable placement of chessmen on the given chessboard.

Input

The first line contains two space-separated integers n and m(1 ≤ n, m ≤ 100).

Each of the next n lines contains a string of m characters: the j-th character of the i-th string is either "." or "-". A "." means that the corresponding cell (in the i-th row and the j-th column) is good, while a "-" means it is bad.

Output

Output must contain n lines, each line must contain a string of m characters. The j-th character of the i-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.

If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.

Sample Input

Input

1 1
.

Output

Input

2 2
..
..

Output

BW
WB

Input

3 3
.-.
---
--.

Output

B-B
---
--B

Hint

In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.

In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.

In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.

先暴力出一個相鄰都不一樣顏色的表，'.'根據表輸出結果，'-'直接輸出'-'。

#include<stdio.h>
#define maxn 109
int map[maxn][maxn];

void init()
{
    int i,j,t;
    for(i = 0;i<maxn;i++)
    {
        if(i%2) t=1;
        else t=0;
        for(j=0;j<maxn;j++)
        {
            map[i][j]=t;
            t=!t;    
        }
    }
}
int main()
{
    int i,j;
    init();
    int n,m;
    char ss;
    while(~scanf("%d %d",&n,&m))
    for(i = 0;i<n;i++)
    {
      getchar();
        for(j=0;j<m;j++)
        {
        
        ss=getchar();
        if(ss == '-')
            printf("-");
        else
            printf("%c",map[i][j]?'B':'W');
        }
        printf("\n");
    }
    return 0;
}

View Code

DZY Loves Chemistry

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Appoint description:

Description

DZY loves chemistry, and he enjoys mixing chemicals.

DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.

Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.

Find the maximum possible danger after pouring all the chemicals one by one in optimal order.

Input

The first line contains two space-separated integers n and m $\text{[math]}$ .

Each of the next m lines contains two space-separated integers x_i and y_i(1 ≤ x_i < y_i ≤ n). These integers mean that the chemical x_iwill react with the chemical y_i. Each pair of chemicals will appear at most once in the input.

Consider all the chemicals numbered from 1 to n in some order.

Output

Print a single integer — the maximum possible danger.

Sample Input

Input

1 0

Output

Input

2 1
1 2

Output

Input

3 2
1 2
2 3

Output

4
題目大意：有n種化學物質跟m種反應關係，試管初始的危險係數爲1，每次往試管中添加物質，若與前面的物質發生了化學反應那麼危險係數乘以2，求最大危險係數。
解題思路：用並查集找出各個集合，每一個結合的最大發生反應次數爲它物質的個數-1。換種思路，就是求s=物質的總數-集合個數,結果爲2^t。

 1 #include<stdio.h>
 2 #define maxn 105
 3 typedef long long LL;
 4 int fa[maxn];
 5 int find(int x)
 6 {
 7     while(fa[x] != x)
 8     {
 9         x = fa[x];
10     }
11     return x;
12 }
13 void merge(int x,int y)
14 {
15     int a = find(x);
16     int b = find(y);
17     if(a != b)
18         fa[a] = b;
19 }
20 
21 int main()
22 {
23     int n,m,i,a,b;
24     scanf("%d%d",&n,&m);
25     for(i=0;i<maxn;i++)
26         fa[i] = i;
27     for(i = 0;i<m;i++)
28     {
29         scanf("%d%d",&a,&b);
30         merge(a,b);
31     }
32     int t=0;
33     LL ans =1;
34     for(i = 1;i<=n;i++)
35     {
36         if(fa[i] == i)t++;
37     }
38     ans = ans<<(n-t);
39     printf("%lld\n",ans);
40     return 0;
41 }

View Code

DZY Loves Modification

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Appoint description:

Description

As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations.

Each modification is one of the following:

Pick some row of the matrix and decrease each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.

DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.

Input

The first line contains four space-separated integers n, m, k and p(1 ≤ n, m ≤ 10³; 1 ≤ k ≤ 10⁶; 1 ≤ p ≤ 100).

Then n lines follow. Each of them contains m integers representing a_ij (1 ≤ a_ij ≤ 10³) — the elements of the current row of the matrix.

Output

Output a single integer — the maximum possible total pleasure value DZY could get.

Sample Input

Input

2 2 2 2
1 3
2 4

Output

Input

2 2 5 2
1 3
2 4

Output

Hint

For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:


1 1
0 0

For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:


-3 -3
-2 -2

題目大意：一個n*m的矩陣有k次操做,每次操做選一行或一列同時減去p操做的價值爲這一行或一列沒操做以前的和,求最大的價值。
解題思路：一共操做k次，能夠設行操做i次,列操做k-i次，每次都是最優操做即選那一行或那一列的和最大的，對i從0到k枚舉結果。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <queue>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 typedef __int64 LL;
 8 const int maxn=1005;
 9 const int maxm=1000005;
10 int map[maxn][maxn];
11 LL C[maxm],R[maxm];
12 priority_queue<LL>sc,sr;
13 LL max(LL a,LL b){return a>b?a:b;}
14 
15 int main()
16 {
17     int i,j,n,m,k,p;
18     LL ans,s;
19     while(~scanf("%d %d %d %d",&n,&m,&k,&p))
20     {
21         while(!sc.empty()) sc.pop();
22         while(!sr.empty()) sr.pop();
23         memset(C,0,sizeof(C));
24         memset(R,0,sizeof(R));
25         for(i=0;i<n;i++)
26         for(j=0;j<m;j++)
27             scanf("%d",&map[i][j]);
28         for(i=0;i<n;i++)
29         {
30             s=0;
31             for(j=0;j<m;j++)
32                 s+=map[i][j];
33             sc.push(s);
34         }
35         for(j=0;j<m;j++)
36         {
37             s=0;
38             for(i=0;i<n;i++)
39                 s+=map[i][j];
40             sr.push(s);
41         }
42         for(i=1;i<=k;i++)
43         {
44             s=sc.top();sc.pop();
45             C[i]=C[i-1]+s;
46             s=s-p*m;
47             sc.push(s);
48         }
49         for(i=1;i<=k;i++)
50         {
51             s=sr.top();sr.pop();
52             R[i]=R[i-1]+s;
53             s=s-p*n;
54             sr.push(s);
55         }
56         ans=-1e18;
57         for(i=0;i<=k;i++)
58             ans=max(ans,C[i]+R[k-i]-(LL)i*(k-i)*p);
59         printf("%I64d\n",ans);
60     }
61     return 0;
62 }

View Code

Artem and Array

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Appoint description:

Description

Artem has an array of n positive integers. Artem decided to play with it. The game consists of n moves. Each move goes like this. Artem chooses some element of the array and removes it. For that, he gets min(a, b) points, where a and b are numbers that were adjacent with the removed number. If the number doesn't have an adjacent number to the left or right, Artem doesn't get any points.

After the element is removed, the two parts of the array glue together resulting in the new array that Artem continues playing with. Borya wondered what maximum total number of points Artem can get as he plays this game.

Input

The first line contains a single integer n(1 ≤ n ≤ 5·10⁵) — the number of elements in the array. The next line contains n integers a_i(1 ≤ a_i ≤ 10⁶) — the values of the array elements.

Output

In a single line print a single integer — the maximum number of points Artem can get.

Sample Input

Input

5
3 1 5 2 6

Output

Input

5
1 2 3 4 5

Output

Input

5
1 100 101 100 1

Output

102

題目大意：給一個n的整數序列，每次刪除一個數這次操做的價值爲它左右兩個數間的較小值，若左邊或右邊沒有數那這次操做的價值爲0，求最大價值。
解題思路：本着這波不虧的貪心思想，先從那些刪了這個數不虧的開始，即左右兩個數均大於等於它。當這樣的好事作完後，新序列有三種狀況（1）單調遞增（2）單調遞減（3）先增後減。這種狀況價值最大的狀況爲序列和減去最大的兩個數（絕對不坑）。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 typedef long long LL;
 7 inline int min(int a,int b){ return a<b?a:b;}
 8 const int maxn=500005;
 9 int n,f[maxn];
10 
11 int main()
12 {
13     int i,top,x;
14     LL ans;
15     while(~scanf("%d",&n))
16     {
17         ans=0;top=0;
18         for(i=0;i<n;i++)
19         {
20             scanf("%d",&x);  
21             while(top>1&&f[top-2]>=f[top-1]&&x>=f[top-1])  
22             {  
23                 ans+=min(x,f[top-2]);  
24                 top--;  
25             }  
26             f[top++]=x;  
27         }
28         sort(f,f+top);
29         for(i=0;i<top-2;i++) ans+=f[i];
30         printf("%lld\n",ans);
31     }
32     return 0;
33 }

View Code