Codeforces Round # FF (Div. 1)-A, B, C

Codeforces Round # FF (Div. 1)-A, B, C

A: DZY Loves Sequences

At first, I read the wrong question .. Sad.

The question is very simple, and the practice is very simple. DP.

Dp [I] [0]: The length obtained from the current position without changing the number.

Dp [I] [1]: to the current position, the length of a number is changed.

However, you need to forward and reverse the request.

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Using namespace std; # define maxn 111_int dp [maxn] [3]; int num [maxn]; int a [maxn]; int n; void dos (int maxx) {memset (dp, 0, sizeof (dp); memset (num,-1, sizeof (num); for (int I = n; I> = 1; I --) {if (a [I] = a [I + 1]) {if (dp [I] [1]
     
      
A [I-1]) {dp [I] [0] = dp [I-1] [0] + 1;} else {dp [I] [0] = 1 ;} dp [I] [1] = dp [I] [0]; num [I] = a [I]; if (a [I]> num [I-1]) {if (dp [I] [1]
      
        B: DZY Loves Modification
       

       
We can find that the size order of a vertical row remains unchanged, but every vertical row drops by p.
       
So we can enumerate x horizontal lines and y vertical lines.
       

       
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Using namespace std; # define maxn 1100 # define LL _ int64int mp [maxn] [maxn]; int hh [maxn]; int ll [maxn]; LL ph [1100000]; LL pl [1100000]; priority_queue
             
               Que; int n, m, k, p; void chu () {ph [0] = pl [0] = 0; while (! Que. empty () que. pop (); for (int I = 1; I <= n; I ++) {que. push (hh [I]) ;}for (int I = 1; I <= k; I ++) {int x = que. top (); que. pop (); ph [I] = ph [I-1] + (LL) x; x = x-p * m; que. push (x) ;}while (! Que. empty () que. pop (); for (int I = 1; I <= m; I ++) {que. push (ll [I]) ;}for (int I = 1; I <= k; I ++) {int x = que. top (); que. pop (); pl [I] = pl [I-1] + (LL) x; x = x-p * n; que. push (x) ;}} int main () {while (~ Scanf ("% d", & n, & m, & k, & p) {memset (hh, 0, sizeof (hh )); memset (ll, 0, sizeof (ll); for (int I = 1; I <= n; I ++) {for (int j = 1; j <= m; j ++) {scanf ("% d", & mp [I] [j]); hh [I] + = mp [I] [j]; ll [j] + = mp [I] [j] ;}} chu (); LL ans = pl [k]; for (int I = 1; I <= k; I ++) {LL x = (LL) I * (LL) (k-I); x = (LL) x * (LL) p; ans = max (ans, pl [k-I] + ph [I]-x );} cout
              
               
There are two main properties:
               
1. The addition of two Fibonacci series is still a Fibonacci series.
               
2. According to the first two items of the Fibonacci series, the Fibonacci numbers at any position can be obtained in O (1) time, and the combination of the Fibonacci series of any length.
               
The rest is a simple question of range summation.
               

               
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                      #pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define mem(a,b) (memset(a),b,sizeof(a))#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now int l,int r,int rt#define INF 99999999#define LL __int64#define mod 1000000009#define eps 1e-6#define zero(x) (fabs(x)
                       
                        r||rr
                        
                         =r) { f1[rt]+=fib[l-ll+1]; f2[rt]+=fib[l-ll+2]; sum[rt]+=suan(fib[l-ll+1],fib[l-ll+2],r-l+1); sum[rt]=(sum[rt]+mod)%mod; f1[rt]=f1[rt]%mod; f2[rt]=f2[rt]%mod; return; } push_down(now); updata(ll,rr,lson); updata(ll,rr,rson); push_up(now);}LL query(int ll,int rr,int_now){ if(ll>r||rr
                         
                          =r)return sum[rt]; push_down(now); return (query(ll,rr,rson)+query(ll,rr,lson))%mod;}int main(){ fib[1]=1;fib[2]=1; for(int i=3;i