noip simulation tests 13

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T1: Matrix Games

　　Just look at the title look ignorant, I feel a little have no idea ......

　　And then began to look at the subject, it seems to identify problems described hiding something? ? ?

　　Idle, bored crazy feeling topic describes a very simple thing, as if to suggest something!

　　So put title equation write down - $ (i-1) * m + j $

　　Then the answer to this formula represented - $ \ sum_ {i = 1} ^ nR_i \ sum_ {j = 1} ^ mS_j [(i-1) * m + j] $

　　Then the item with a $ I $ can be ripped out of $ O (n) $ done

　　

　　so，code

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<vector>
 7 #define ll long long
 8 using namespace std;
 9 const int MAXN=1000005,MAXK=100005,D=1e9+7;
10 int n,m,k;
11 ll sums,sumr,s[MAXN],r[MAXN],ans,tmp,sum;
12 char opt[5];
13 int main() {
14     scanf("%d%d%d",&n,&m,&k);
15     for(int i=1;i<=m;i++) s[i]=1;
16     for(int i=1;i<=n;i++) r[i]=1;
17     for(int i=1;i<=k;i++) {
18         ll x,y;
19         scanf("%s%lld%lld",opt,&x,&y);
20         if(opt[0]=='R') r[x]=r[x]*y%D;
21         else s[x]=s[x]*y%D;
22     }
23     for(int i=1;i<=m;i++) sums=(sums+s[i])%D;
24     for(int i=1;i<=m;i++) tmp=(tmp+s[i]*i)%D;
25     for(int i=1;i<=n;i++) ans=(ans+(tmp+(ll)(i-1)*m%D*sums)%D*r[i]%D)%D;
26     printf("%lld\n",(ans%D+D)%D);
27     return 0;
28 }

t1 Code

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T2: Hopscotch

 　　See the range of k matrix multiplication first thought, but looked n and m or forget

　　Found that most changes will only affect the previous operation three points, so even violent side run,

　　Also found that if a graph is built of n points, n edges directed graph, it shall be a ring inward of the tree

　　It is easy to think of the time when a lot of k is always around the ring, so you can take it to the length of the ring mold, the complexity of the almost

　　

 

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T3: beautiful sequence

 

 

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