几个小结论: 1.路径长度=i+j-1; 2.最简单的走法是先横走再竖着走或者先竖着走再横着走 #include<cstdio> #include<cstring> using namespace std; void print(int x,int y) { ; i<=y; i++) printf("(1,%d) ",i); ; i<=x; i++) printf("(%d,%d) ",i,y); printf("\n&…

B. Inna and New Matrix of Candies time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Can…

B. Inna and New Matrix of Candies time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Can…

B. Inna and New Matrix of Candies time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Can…

本题的主要算法就是区间更新和区间求和: 可以用线段树和树状数组来做: 感觉线段树写的太麻烦了,看到官方题解上说可以用树状数组做,觉得很神奇,以前用过的树状数组都是单点维护,区间求和的: 其实树状数组还可以区间维护,单点求值:和区间维护,区间求和的: 详情请见博客. #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #define maxn 4000010 #define…

树状数组仅仅能实现线段树区间改动和区间查询的功能,能够取代不须要lazy tag的线段树.且代码量和常数较小 首先定义一个数组 int c[N]; 并清空 memset(c, 0, sizeof c); 1.单点改动 : c[x] += y; 相应的函数是 change(x, y); 2.求前缀和 :  相应的函数是 int sum(x) 两种操作的复杂度都是O(logn) 模板例如以下: int c[N], maxn; inline int Lowbit(int x){return x&(-x…

题目链接:点击打开链接 题意:给定n*m的二维平面 w个操作 int mp[n][m] = { 0 }; 1.0 (x1,y1) (x2,y2) value for i : x1 to x2 for j : y1 to y2 mp[i][j] += value; 2.1 (x1, y1) (x2 y2) ans1 = 纵坐标在 y1,y2间的总数 ans2 = 横坐标不在x1,x2间的总数 puts(ans1-ans2); more format: for i : 1 to n for j :…

http://codeforces.com/problemset/problem/400/C 题意:给你一个n*m的矩阵,然后在矩阵中有p个糖果,给你每个糖果的初始位置,然后经过x次顺时针反转,y次旋转,z次逆时针反转,问最后每个糖果的位置. 思路:推出顺时针反转.旋转.逆时针反转的坐标的变化即可. #include <cstdio> #include <cstring> #include <cmath> #include <iostream> #defin…

题目链接:http://codeforces.com/problemset/problem/400/C 题目意思:给出一个n行m列的矩阵,问经过 x 次clockwise,y 次 horizontal rotate 和z次counterclockwise 之后,原来在n行m列的矩阵的坐标去到哪个位置. 题目意思很容易看懂.易知,对于clockwise,counterclockwise的次数,mod 4 == 0 相当于没有改变!而对于 horizontal rotate,mod 2 == 0 也…

#include <iostream> #include <vector> #include <string> #include <algorithm> #include <set> using namespace std; int main(){ int n,m; cin >> n >> m; set<int> distance; bool flag = true; ; i < n; ++ i){ st…