[C++知识库]刷题 3/11

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;

const int fx[] = {0, -2, -1, 1, 2, 2, 1, -1, -2};
const int fy[] = {0, 1, 2, 2, 1, -1, -2, -2, -1};//Use two one-dimensional arrays to determine where Marco can walk

int bx, by, hx, hy;
ll f[40][40];
bool s[40][40]; //Determine if there are horses at this point
int main(){
    scanf("%d%d%d%d", &bx, &by, &hx, &hy);
    bx += 2; by += 2; hx += 2; hy += 2;//To prevent cross-border
    f[2][1] = 1;//Initialization
    s[hx][hy] = 1;//tag
    for(int i = 1; i <= 8; i++) s[hx + fx[i]][hy + fy[i]] = 1;
    for(int i = 2; i <= bx; i++){
        for(int j = 2; j <= by; j++){
            if(s[i][j]) continue; //If you're stopped by a horse, jump over it
            f[i][j] = f[i - 1][j] + f[i][j - 1];//State transition equation
        }
    }
    printf("%lld\n", f[bx][by]);
    return 0;
} 

对于状态的变化，按照现态和次态，按照逻辑顺序排序，不容易漏掉/重复。
找出状态转化方程，便于动态规划的分析

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int fx[] = {0, -2, -1, 1, 2, 2, 1, -1, -2};
const int fy[] = {0, 1, 2, 2, 1, -1, -2, -2, -1};//Use two one-dimensional arrays to determine where Marco can walk

int bx, by, hx, hy;
ll f[40][40];
bool s[40][40]; //Determine if there are horses at this point
int main(){
    scanf("%d%d%d%d", &bx, &by, &hx, &hy);
    bx += 2; by += 2; hx += 2; hy += 2;//To prevent cross-border
    f[2][1] = 1;//Initialization
    s[hx][hy] = 1;//tag
    for(int i = 1; i <= 8; i++) s[hx + fx[i]][hy + fy[i]] = 1;
    for(int i = 2; i <= bx; i++){
        for(int j = 2; j <= by; j++){
            if(s[i][j]) continue; //If you're stopped by a horse, jump over it
            f[i][j] = f[i - 1][j] + f[i][j - 1];//State transition equation
        }
    }
    printf("%lld\n", f[bx][by]);
    return 0;
} 

C++万能头文件

#include <bits/stdc++.h>

遇到不知道输入多少组数据的题目->while

C语言

While(scanf(“%d”,&n)!=EOF){
}

C++

While(cin>>a>>b){
}