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。
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 5for 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 namespace std int maxn="1000010;int" main string s while>>s) { int n= s.size(); int cnt = s[0]-'0'; for(int i = 1; i <br> C:给你一些数,你取了一个数那么比这个数大1,和小1的数字就会被删掉。问你最大能取到的数的和。 <p></p> <p>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。</p> <p></p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> C. Boredom </p> <p class="sycode"> </p> <p class="sycode"> time limit per test </p> 1 second <p class="sycode"> </p> <p class="sycode"> memory limit per test </p> 256 megabytes <p class="sycode"> </p> <p class="sycode"> input </p> standard input <p class="sycode"> </p> <p class="sycode"> output </p> standard output <p class="sycode"> </p> <p> 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.</p> <p> 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.</p> <p> Alex is a perfectionist, so he decided to get as many points as possible. Help him.</p> <p class="sycode"> </p> <p class="sycode"> Input </p> <p> The first line contains integer n (1?≤?n?≤?105) that shows how many numbers are in Alex's sequence.</p> <p> The second line contains n integers a1, a2, ..., an (1?≤?ai?≤?105).</p> <p class="sycode"> </p> <p class="sycode"> Output </p> <p> Print a single integer ? the maximum number of points that Alex can earn.</p> <p class="sycode"> </p> <p class="sycode"> Sample test(s) </p> <p class="sycode"> </p> <p class="sycode"> </p> <p class="sycode"> input </p> <pre style="代码" class="precsshei">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 namespace std int maxn="1000010;LL" vis dp main n while>>n) { int x; memset(vis, 0, sizeof(vis)); memset(dp, 0, sizeof(dp)); for(int i = 0; i <br> <br> <p></p> </eps></set></map></ctime></stack></cmath></queue></string></stdio.h></iomanip></string.h></stdlib.h></iostream></algorithm>