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        codeforces260div2A,B,C_html/css

        來源:懂視網(wǎng) 責(zé)編:小采 時間:2020-11-27 15:54:33
        文檔

        codeforces260div2A,B,C_html/css

        codeforces260div2A,B,C_html/css_WEB-ITnose:A:水題,結(jié)構(gòu)體排序后,看兩個數(shù)組的是否序列相同。 B:分別寫出來1,2,3,4,的n次方對5取余。你會發(fā)現(xiàn)和對5取余有一個循環(huán)節(jié)。如果%4 = 0,輸出4,否則輸出0. 寫一個大數(shù)取余就過了。 B. Fedya and Maths time limit per test
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        導(dǎo)讀codeforces260div2A,B,C_html/css_WEB-ITnose:A:水題,結(jié)構(gòu)體排序后,看兩個數(shù)組的是否序列相同。 B:分別寫出來1,2,3,4,的n次方對5取余。你會發(fā)現(xiàn)和對5取余有一個循環(huán)節(jié)。如果%4 = 0,輸出4,否則輸出0. 寫一個大數(shù)取余就過了。 B. Fedya and Maths time limit per test

        A:水題,結(jié)構(gòu)體排序后,看兩個數(shù)組的是否序列相同。

        B:分別寫出來1,2,3,4,的n次方對5取余。你會發(fā)現(xiàn)和對5取余有一個循環(huán)節(jié)。如果%4 = 0,輸出4,否則輸出0.

        寫一個大數(shù)取余就過了。

        B. Fedya and Maths

        time limit per test

        1 second

        memory limit per test

        256 megabytes

        input

        standard input

        output

        standard output

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

        (1n?+?2n?+?3n?+?4n) mod 5

        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 #include #include #include #include #include #include #include #include #include #include #include #include #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)>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<

        C:給你一些數(shù),你取了一個數(shù)那么比這個數(shù)大1,和小1的數(shù)字就會被刪掉。問你最大能取到的數(shù)的和。

        先根據(jù)數(shù)字進(jìn)行哈希,然后線性的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代表取這個數(shù),0代表不取。注意數(shù)據(jù)類型要用long long。

        C. Boredom

        time limit per test

        1 second

        memory limit per test

        256 megabytes

        input

        standard input

        output

        standard 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 #include #include #include #include #include #include #include #include #include #include #include #include #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)>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<

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        文檔

        codeforces260div2A,B,C_html/css

        codeforces260div2A,B,C_html/css_WEB-ITnose:A:水題,結(jié)構(gòu)體排序后,看兩個數(shù)組的是否序列相同。 B:分別寫出來1,2,3,4,的n次方對5取余。你會發(fā)現(xiàn)和對5取余有一個循環(huán)節(jié)。如果%4 = 0,輸出4,否則輸出0. 寫一個大數(shù)取余就過了。 B. Fedya and Maths time limit per test
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        標(biāo)簽: it cc 2a
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