JinweiGuo

输入与输出 输入 数组输入: 可以提前定义 a = [0 for i in range(1000)] 或者 arr = list(map(int,input().split())) 输出 import math pi = math.pi p = -1*pi print(pi) print('%.4f' % pi) # 2,前面补充前导0表示月份 a = 2 # 2月输出02 b = 11 # 11月输出11 print('%02d' % a) print('%02d' % b) print('%d %d'

却道一句当时只道是寻常 python国赛复盘 我是个c++选手，python就国赛前学了半个下午加晚上，省赛前学了半天。所以填空题部分还是选择了，用c++编程，机房提供了DEV还是很不错的 print(200/8) # 答案是25 #include <iostream> #include <cstring> #include <algorithm> #include <stdio.h> using namespace std; typedef long

牛客挑战赛47 A **多维的更相减损术得gcd(x,y,z) = gcd(x,y-x,z-y) ** 则gcd(a1+k,a2+k,a3+k…,an+k) = gcd(a1+k,a2-a1,a3-a2,a4-a3…,an-an-1) 提前排好序 #include <iostream> #include <algorithm> using namespace std; const int maxn = 1e5+10; long long a[maxn],n; int main() {

Problem A Problem Description Mo’s algorithm can solve some difficult data structure problems. Like this: You are given an array a1,a2,…,an and m queries. In each query, we want to get the mode number(the value that appears most often in a set of data) of