摘要：#include#includeusing namespace std;void printMonth(int year, int month);void printMonthTitle(int year, int month);void printMonthName(int month);void... 阅读全文

摘要：You guys must have seen nested list (嵌套链表) such as [1, 2, 3, [4, 5, [6, 7]], 8, [9, 0]].So your task is to improve DouList with nested feature. You ca... 阅读全文

摘要：Description有一天, PY找到了一张藏宝图.这张地图被划分成2^n * 2^n个格子.左上角的格子的坐标为(0, 0)而右下角格子的坐标为(2^n-1, 2^n-1).如下:(0, 0)(0, 1)……(0, 2^n-1)……………………(2^n-2, 0)(2^n-2,1)……(2^n-... 阅读全文

摘要：对于两个整数G和L（1#include#includeint main(){int d[10000];int T, G, L, ans, i, num; scanf("%d\n",&T);while(T--){ memset(d,0,sizeof(d)); num =0; scanf("%d %d"... 阅读全文

摘要：Description概念1：梅森数，当一个正整数可以表示成2p-1形式，且p为素数时，2p-1即称为梅森数概念2：在自然数中，只有1和它本身两个约数的数叫做素数。一个梅森数被称为梅森素数当且仅当它是一个素数。现给定一个正整数p（pint prime(int x){longlong i;if(x <... 阅读全文

摘要：You'll be given two intergers. The number of the digits of each is from 1 to 200, inclusive. And what you need to do is to culcalate the product of th... 阅读全文

摘要：Given a b and p, output (a^b) % p (2#include#includeint quick_power(int a,int b,int p){int temp;if(b ==0){return1;} temp = quick_power(a, b /2, p);if(... 阅读全文

摘要：ConstraintsTime Limit: 1 secs, Memory Limit: 8 MBDescriptionInput A and B, output A+BInputInput two values, A and B.(0#includevoid init(int*arr,int*le... 阅读全文

摘要：Description桌上有一叠牌，从第一张牌（即位于顶面的牌）开始从上往下依次编号为1~n。当至少还剩两张牌时进行以下操作：把第一张牌扔掉，然后把新的第一张放到整叠牌的最后。输入n,输出每次扔掉的牌，以及最后剩下的牌。Input第一行为一个整数t（0int main(){int test, n, ... 阅读全文

摘要：F[0]=0;F[1]=1;F[n]=F[n-1]+F[n-2], for n>1给出n (0#include// 矩阵快速幂void xx(longlong a[][3],longlong b[][3],longlong m){longlong tmp[3][3]; tmp[1][1]= tmp[... 阅读全文

摘要：// List.h#ifndef SSCPP2014_DOULIST_A_H#define SSCPP2014_DOULIST_A_H#includestructDouListNode{int elem;DouListNode*prev,*next;DouListNode(int e =0,DouListNode*p =0,DouListNode*n =0){ elem = e; prev = p; next = n;}};classDouList{private:DouListNode*m_head,*m_tail;public:DouList();DouList(constDouList& 阅读全文