哇，果然是一曝十寒啊，这么多天没总结了。
再次整理了一下同余模问题：
附上简单入门题：
http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=781
A Math Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1867 Accepted Submission(s): 519
Problem Description
You are given a positive integer n, please count how many positive integers k satisfy kk≤n.
Input
There are no more than 50 test cases.
Each case only contains a positivse integer n in a line.
1≤n≤1018
Output
For each test case, output an integer indicates the number of positive integers k satisfy kk≤n in a line.
Sample Input
1
4
Sample Output
1
2
#include
#include
#define LL long long;
LL kpow(int x, int n){
LL res = 1;
while(n >= 0){
if(n & 1) res = res * x;
x = x * x;
n >>= 1;
}
return res;
}
int main(){
LL n;
while(~scanf("%lld",&n)){
for(int k = 15; k >= 1; k--){
if(kpow(k,k) <= n){
printf("%d\n",k);
break;
}
}
}
return 0;
}
今天人工智能考到A算法，错的惨兮兮
回来补补课
POJ 2243 一道A入门题
Knight Moves
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14679 Accepted: 8226
Description
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.
#include
#include
#include
#include
#include
#include
using namespace std;
struct knight{
int x,y,step;
int g,h,f;
bool operator < (const knight &k) const{
return f > k.f;
}
}k;
bool visited[8][8];
int x2,y2,ans;
int dir[8][2] = {{-2,-1},{-2,1},{2,-1},{2,1},{-1,-2},{-1,2},{1,-2},{1,2}};
priority_queue que;
bool in (const knight &a){
if(a.x < 0 || a.y < 0 || a.x >= 8 || a.y >= 8){
return false;
}
return true;
}
//曼哈顿估价函数
int Heuristic(const knight & a){
return (abs(a.x - x2) + abs(a.y - y2)) * 10;
}
void Astar(){
knight t,s;
while(!que.empty()){
t = que.top();
que.pop();
visited[t.x][t.y] = true;
if(t.x == x2 && t.y == y2){
ans = t.step;
break;
}
for(int i = 0; i < 8; i++){
s.x = t.x + dir[i][0];
s.y = t.y + dir[i][1];
if(in(s) && !visited[s.x][s.y]){
s.g = t.g + 23;
s.h = Heuristic(s);
s.f = s.g + s.h;
s.step = t.step + 1;
que.push(s);
}
}
}
}
int main(){
char line[5];
int x1,y1;
while(gets(line)){
x1 = line[0] - 'a';
y1 = line[1] - '1';
x2 = line[3] - 'a';
y2 = line[4] - '1';
memset(visited,false,sizeof(visited));
k.x = x1;
k.y = y1;
k.g = k.step = 0;
k.h = Heuristic(k);
k.f = k.g + k.h;
while(!que.empty()) que.pop();
que.push(k);
Astar();
printf("To get from %c%c to %c%c takes %d knight moves.\n",line[0],line[1],line[2],line[3],ans);
}
return 0;
}
另外这题还可以用双向BFS解决，待补坑。。。
再挖一坑，最近碰到线段树这一新的数据结构，一道入门题练手 HDU 1166，挖坑待补。。。
四子连棋问题
BFS + 数码状态表示 + hash判重
#include
#include