codeforces 260 div2 A,B,C_html/css_WEB-ITnose

a:水题，结构体排序后，看两个数组的是否序列相同。

B：分别写出来1，2，3，4，的n次方对5取余。你会发现和对5取余有一个循环节。如果%4 = 0，输出4，否则输出0.

写一个大数取余就过了。

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B. Fedya and Maths

time limit per test

memory limit per test

input

output

Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:

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for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

Input

The single line contains a single integer n (0?≤?n?≤?10105). The number doesn't contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Sample test(s)

input

output

input

124356983594583453458888889

output

Note

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <ctime>#include <map>#include <set>#define eps 1e-9///#define M 1000100///#define LL __int64#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)using namespace std;const int maxn = 1000010;int main(){    string s;    while(cin >>s)    {        int n= s.size();        int cnt = s[0]-'0';        for(int i = 1; i < n; i++)        {            cnt %= 4;            cnt = (cnt*10+(s[i]-'0'))%4;        }        if(cnt%4 == 0)            cout<<4<<endl;        else cout<<0<<endl;    }}

先根据数字进行哈希，然后线性的dp一遍，dp[i][1] = ma(dp[i-2][0], dp[i-2][1]) + vis[i]*i,dp[i][0] = max(dp[i-1][0], dp[i-1][1]).1代表取这个数，0代表不取。注意数据类型要用long long。

C. Boredom

time limit per test

memory limit per test

input

output

Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.

Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak?+?1 and ak?-?1 also must be deleted from the sequence. That step brings ak points to the player.

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

Input

The first line contains integer n (1?≤?n?≤?105) that shows how many numbers are in Alex's sequence.

The second line contains n integers a1, a2, ..., an (1?≤?ai?≤?105).

Output

Print a single integer ? the maximum number of points that Alex can earn.

Sample test(s)

input

21 2

output

input

31 2 3

output

input

91 2 1 3 2 2 2 2 3

output

10

Note

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2,?2,?2,?2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <ctime>#include <map>#include <set>#define eps 1e-9///#define M 1000100///#define LL __int64#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)using namespace std;const int maxn = 1000010;LL vis[maxn];LL dp[maxn][2];int main(){    int n;    while(cin >>n)    {        int x;        memset(vis, 0, sizeof(vis));        memset(dp, 0, sizeof(dp));        for(int i = 0; i < n; i++)        {            scanf("%d",&x);            vis[x] ++;        }        dp[1][1] = vis[1];        dp[2][1] = vis[2]*2;        dp[2][0] = dp[1][1];        for(int i = 3; i <= maxn-10; i++)        {            dp[i][1] = max(dp[i-2][0], dp[i-2][1])+vis[i]*i;            dp[i][0] = max(dp[i-1][0], dp[i-1][1]);        }        cout<<max(dp[maxn-10][0], dp[maxn-10][1])<<endl;    }    return 0;}