"Analog 8.29" chinese (nature) · physics · chemistry (probability expectation)

T1  chinese

According to his question i * f [i] we tend to think of the answer is in fact the contribution of each scenario for each point is 1 plus and

We can change the problem, the contribution of each point in all programs

And then actually asking is 1-k values, the number of the program and then take a

In fact do so is each point pulled out, although each program may contain a number of possible points, but every time we consider just one point contribution, so the correct

ps: not modulo index

Code is very short

 1 #include<bits/stdc++.h>
 2 #define int long long
 3 using namespace std;
 4 int n,m,K;
 5 int mod=1e9+7;
 6 int poww(int x,int y)
 7 {
 8      int ans=1;
 9      while(y)
10      {
11          if(y&1)ans=ans*x%mod;
12          x=x*x%mod;
13          y>>=1;
14      }
15      return ans%mod;
16 }
17 int sum=0;
18 signed main()
19 {
20     scanf("%lld%lld%lld",&n,&m,&K);        
21     int zz=poww(K,n*m-n-m+1);
22     int two=(n-1)+(m-1);
23     for(int k=2;k<=K;++k)
24     {
25         int xx=poww(k-1,two);
26         sum=(sum+(xx*zz)%mod)%mod;
27     }
28     printf("%lld\n",sum*n%mod*m%mod);
29 }

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T2 physics

Positive solution did not play, the cushions

Violence pruning off water

A pruning: the prefix and processing O (1) query, but because the value of each change is small

Just when determining whether to modify modify symbols in the last round of the answer, the answer is the output, otherwise the changes

Pruning two: the right side of each half of the length of the border is the last round of answers

 1 #include<bits/stdc++.h>
 2 #define int long long
 3 #define MAXN 2011
 4 using namespace std;
 5 int n,m;
 6 char a[MAXN][MAXN];
 7 int sum[MAXN][MAXN];
 8 int last_x,last_y,last_len;
 9 int zuo_x,zuo_y;
10 int get_sum1(int x1,int y1,int x2,int y2)
11 {
12     return sum[x2][y2]+sum[x1-1][y1-1]-sum[x2][y1-1]-sum[x1-1][y2];
13 }
14 int work(int mid)
15 {
16     for(int i=1;i<=n-mid+1;++i)
17     {
18         for(int j=1;j<=m-mid+1;++j)
19         {
20             if(get_sum1(i,j,i+mid-1,j+mid-1)==0)
21             {
22                 last_x=i,last_y=j;last_len=mid;
23                 return 1;
24             }
25         }
26     }
27     return 0;
28 }
29 void second_divied()
30 {          
31     int l=1;int r=last_len;
32     while(l+1<r)
33     {    
34           int mid=(l+r)>>1;
35           if(work(mid))l=mid;
36           else r=mid;   
37     }
38     if(work(r))printf("%lld\n",r);
39     else if(work(l))printf("%lld\n",l);
40 }          
41 int q;int ok=0;
42 void the_yu(int x,int y)
43 {
44      for(int i=x;i<=n;++i)
45      {
46          for(int j=y;j<=m;++j)
47          {
48              sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
49              if(a[i][j]=='-')sum[i][j]++;
50          }
51      }
52 }
53 bool pan(int x,int y)
54 {
55      if(x>=last_x&&y>=last_y&&x<=last_x+last_len-1&&y<=last_y+last_len-1)return 0;
56      return 1;
57 }
58 signed main()
59 {
60     scanf("%lld%lld%lld",&n,&m,&q);last_len=min(n,m);
61     for(int i=1;i<=n;++i)scanf("%s",a[i]+1);
62     zuo_x=0x7fffff;zuo_y=0x7ffff;    
63     for(int i=1;i<=q;++i)
64     {
65        int x,y;
66        scanf("%lld%lld",&x,&y);
67        a[x][y]='-';
68        zuo_x=min(zuo_x,x);zuo_y=min(zuo_y,y);
69        if(i>1&&pan(x,y))
70        {
71            printf("%lld\n",last_len);
72            continue;
73        }
74        else
75        {
76             IF (I == . 1 ) the_yu ( . 1 , . 1 );
 77             the else the_yu (zuo_x, zuo_y);
 78             second_divied ();
 79             zuo_x = 0x7FFFFF ; zuo_y = 0x7FFFFF ;
 80         }
 81      }
 82 }

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T3 chemistry

The probability of expectation, if not done, the goo ....