Codeforces Round #FF (Div. 1)-A,B,C_html/css_WEB-ITnose

A:DZY Loves Sequences

I read the wrong question at first. . sad.

The question is very simple and the method is also very simple. Just DP it.

dp[i][0]: The length obtained without any number change to the current position.

dp[i][1]: Go to the current position, change a number, and get the length

But you need to find it once in the forward direction, and then in the reverse direction.

#include<iostream>#include<stdio.h>#include<algorithm>#include<stdlib.h>#include<string.h>using namespace std;#define maxn 110000int 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 dp else num if>=a[i+1])        {            if(dp[i][1]<dp dp num for i="1;" maxx="max(maxx,dp[i][0]);" cout main while scanf memset if>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]<dp dp num if int maxx="-1;" for i="1;" cout dos return>B: DZY Loves Modification <p></p> <p>We can find that when selecting a horizontal row, the size order of the vertical rows remains unchanged, but each vertical row decreases by p . </p> <p>So we can enumerate and select x horizontal rows and y vertical rows. </p> <p></p> <pre name="code" class="sycode">#include<iostream>#include<stdio.h>#include<algorithm>#include<stdlib.h>#include<string.h>#include<queue>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<int>que;int n,m,k,p;void chu(){    ph[0]=pl[0]=0;    while(!que.empty())que.pop();    for(int i=1;iC: DZY Loves Fibonacci Numbers <p></p> <p>Mainly two properties: </p> <p>1, the addition of two Fibonacci numbers It’s still a Fibonacci sequence. </p> <p>2. According to the first two terms of the Fibonacci sequence, the Fibonacci number at any position and the sum of the Fibonacci sequence of any length can be obtained in O(1) time. </p> <p>The rest is a simple interval summation problem. </p> <p></p> <pre name="code" class="sycode">#include<stdio.h>#include<iostream>#include<stdlib.h>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<map>#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,rtr||rr<l if>=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<l cout if>=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<maxn fib int n while creat scanf if else printf return> <br> <br> <br> <p></p> <p><br> </p> </maxn></l></l></map></math.h></vector></algorithm></string.h></stdlib.h></iostream></stdio.h>