F(N) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4078    Accepted Submission(s): 1432 Problem Description Giving the N, can you tell me the answer of F(N)?   Input Each test case contains a…

Problem Description Giving the N, can you tell me the answer of F(N)? Input Each test case contains a single integer N(1<=N<=10^9). The input is terminated by a set starting with N = 0. This set should not be processed. Output For each test case, ou…

找循环节水题.注意余数大于0. /* 2802 */ #include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 4018 #define MOD 2009 __int64 a[MAXN+]; void init() { __int64 tmp; int i, j, k; bool flag; a[]=, a[] = , a[] = ; ; i<=MAXN; ++i) { tmp = i…

数位DP.... F(x) Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 750    Accepted Submission(s): 286 Problem Description For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weig…

题目大意:f[n]-n^3=f[n-2]-(n-1)^3 (n >=3),f[1]=1,f[2]=7,求f[n]. 题目思路:将n^3移到到等式右边化简的到:f[n]=f[n-2]+3n*(n-1)+1: 因为第n项与第n-2项有关,所以知道了f[1]与f[2]的值可以分奇偶打下表,找到循环节为4018. #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #incl…

题目链接 F(N) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4579 Accepted Submission(s): 1610 Problem Description Giving the N, can you tell me the answer of F(N)? Input Each test case contains a si…

题目链接:HDU 2009-4 Programming Contest 分析:具有一定的周期性——4018处理下就可以A了 Sample Input Sample Output AC代码: #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<iostream> #include<queue> #include<map>…

Discription This time I need you to calculate the f(n) . (3<=n<=1000000) f(n)= Gcd(3)+Gcd(4)+…+Gcd(i)+…+Gcd(n). Gcd(n)=gcd(C[n][1],C[n][2],……,C[n][n-1]) C[n][k] means the number of way to choose k things from n some things. gcd(a,b) means the greate…

Giving the N, can you tell me the answer of F(N)? Input Each test case contains a single integer N(1<=N<=10^9). The input is terminated by a set starting with N = 0. This set should not be processed. Output For each test case, output on a line the v…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2802 f[1] = 1; f[2] = 7; f[n] = (f[n-2] - (n-1)*(n-1)*(n-1)+ n*n*n) % 2009, n>2; 给你一个n让你输出f[n];(n<1e9)由于n较大,所以不能直接输出,结果是对2009求余所以一定存在循环节,由于 f(n) 只与 f(n-2) 和 n 有关, 所以我们只需判断 G[n][f[n-2]] 第一次出现和第二次出现的次数即可,…