这道题目可以手工打表,也可以机器打表,千万不能暴力解,会TLE. #include <stdio.h> #define MAXNUM 1000000001 ][]; int main() { int case_n, n; int i, j; ; i<; ++i) { buf[i][] = ; buf[i][] = i; ; j<; ++j) { buf[i][j] = buf[i][j-]*i%; ]) break; buf[i][]++; } } /* for (i=0; i&l…

HDU 1061 题目大意:给定数字n(1<=n<=1,000,000,000),求n^n%10的结果 解题思路:首先n可以很大,直接累积n^n再求模肯定是不可取的, 因为会超出数据范围,即使是long long也无法存储. 因此需要利用 (a*b)%c = (a%c)*(b%c)%c,一直乘下去,即 (a^n)%c = ((a%c)^n)%c; 即每次都对结果取模一次 此外,此题直接使用朴素的O(n)算法会超时,因此需要优化时间复杂度: 一是利用分治法的思想,先算出t = a^(n/2),若…

Rightmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39554    Accepted Submission(s): 14930 Problem Description Given a positive integer N, you should output the most right digit of N…

Problem Description Given a positive integer N, you should output the most right digit of N^N. Input The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each…

找出数学规律 原题: Rightmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6515    Accepted Submission(s): 2454 Problem Description Given a positive integer N, you should output the most right d…

Problem Description Given a positive integer N, you should output the most right digit of N^N. Input The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each…

Problem Description Given a positive integer N, you should output the most right digit of N^N. Input The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.Each…

Rightmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 57430    Accepted Submission(s): 21736 Problem Description Given a positive integer N, you should output the most right digit of N…

二分. #include <stdio.h> #include <math.h> int main() { int case_n; double n, tmp, l, r; int m; scanf("%d", &case_n); while (case_n--) { scanf("%lf", &n); tmp = sqrt(n+n); tmp = ceil(tmp); l = (tmp*tmp-tmp)/; r = (tmp…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…