Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin. this puzzle describes that: gave a and b,how to know…

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…

题目和1061非常相似,几乎可以复用. #include <stdio.h> ][]; int main() { int a, b; int i, j; ; i<; ++i) { buf[i][] = ; buf[i][] = i; ; j<; ++j) { buf[i][j] = buf[i][j-]*i%; ]) break; buf[i][]++; } } while (scanf("%d %d", &a, &b) != EOF) { i…

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1097 分析:简单题,快速幂取模, 由于只要求输出最后一位,所以开始就可以直接mod10. /*A hard puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33036 Accepted Submission(s): 11821 Pr…

trie树.以puzzle做trie树内存不够,从puzzle中直接找串应该会TLE.其实可以将查询组成trie树,离线做.扫描puzzle时注意仅三个方向即可. /* 1857 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector>…

Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integer k(k<10000),to find the minimal nonegative integer…

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…

直接构造矩阵,最上面一行加一排1.高速幂计算矩阵的m次方,统计第一行的和 CRB and Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 133    Accepted Submission(s): 63 Problem Description CRB is now playing Jigsaw Puzzle. There…

dp [ x ] [ y ] [ z ] 表示二进制y所表示的组合对应的之和mod x余数为z的最小数... 如可用的数字为 1 2 3 4...那么 dp [ 7 ] [ 15 ] [ 2 ] = 1234 .... 输入一个数列后..将dp的表做出来..然后O(1)的输出...题目要求是( T + X ) % K =0 可以转化为 T % K = ( K - ( X % K ) ) % K Program: #include<iostream> #include<stdio.h>…