Recursive mastery core:
- Recursion formula (law)
- Recursion boundary (initialization condition)
analysis question meaning: as shown below,
recursion formula: a[n] = a[n-1]+a[n-2];
recursion boundary: a[1]=1 a[2] =2
#include<iostream>
using namespace std;
long long a[100],x,y,z;
int main(){
//昆虫繁衍
cin>>x>>y>>z;
//x月前面
for(int i=1;i<=x;i++){
a[i]=1; //一对成虫
}
//x月后面
for(int i=x+1;i<=z+1;i++){
//递推公式 a[n]=a[n-1]+a[n-x-2]*y
a[i]=a[i-1]+a[i-x-2]*y;
}
cout<<a[z+1];
return 0;
}
Question meaning: How many paths can the pawn take from point A to point B? There are prerequisites: the pawn can only go down or to the right! The horse's position and 8 control points cannot be moved!
Since the pawn keeps going to the right, we know that the P5 point (0,3) is controlled by the horse and cannot move. Then (0,4) cannot move behind P5!
- 递推式：a[i][j]=a[i-1][j]+a[i][j-1];
#include<bits/stdc++.h>
using namespace std;
int a[30][30]; //数组存储这个坐标点的路径数
bool flag[30][30]; //标记这个点是否能走
int b[8][2]={
{
-1,-2},{
1,-2},{
-2,-1},{
2,-1},{
1,2},{
-1,2},{
2,1},{
-2,1}};
int main(){
int n,m,x,y;
cin>>n>>m>>x>>y;
flag[x][y]=true;
//马的坐标x,y 标记true，代表不能走
for(int i=0;i<8;i++){
//马的8个控制点也不能走
int tx = x+b[i][0]; //8个控制点的x坐标
int ty = y+b[i][1]; //8个控制点的y坐标
if(tx>=0 && tx<=n && ty>=0 && ty<=m){
//坐标满足在范围内
flag[tx][ty]=true; //标记true，代表不能走
}
}
for(int i=0;i<=n;i++){
for(int j=0;j<=m;j++){
//遍历整个二维数组
if(flag[i][j]==true){
a[i][j]=0; //这个点的路径为0
}
else if (i==0 && j==0){
a[i][j]=1; //地点位置路径数为1
}
else if(i==0 && j!=0){
//单行(0,1) (0,2)的路径数由前面的位置决定！
a[i][j] = a[i][j-1];
}
else if(j==0 && i!=0){
//单列(1,0) (2,0)的路径数由前面的位置决定！
a[i][j] = a[i-1][j];
}
else{
//递推规律
a[i][j]=a[i-1][j]+a[i][j-1];
}
}
}
cout<<a[n][m];
return 0;
}
This question is relatively simple. You need to pay attention to taking the modulo 1000 of the ath number. It must be implemented when recursive, otherwise an error will be reported in the output because the result of the operation is too large! ! !
Recursive formula: f[i] = (f[i-1]+f[i-2])%1000;
#include<bits/stdc++.h>
using namespace std;
long long f[1000002];
int main(){
f[1]=f[2]=1;
int n,a;
cin>>n;
for(int i=3;i<=1000000;i++){
//斐波那契数列1-100000的值
f[i] = (f[i-1]+f[i-2])%1000; //递推式
}
for(int i=1;i<=n;i++){
//根据题目条件，n行，每一行有第a个数列，输出第a个斐波那契数列值
cin>>a;
cout<<f[a]<<endl;
}
return 0;
}
This question is not difficult, as long as you know the recursion formula and the recursion boundary!
It is obvious that the recurrence formula is: f[i] = (2*f[i-1]+f[i-2])%32767
#include<bits/stdc++.h>
using namespace std;
long long f[1000009];
int main(){
f[1]=1;
f[2]=2;
int n,a;
for(int i=3;i<=1000000;i++){
f[i] = (2*f[i-1]+f[i-2])%32767;
}
cin>>n;
for(int i=1;i<=n;i++){
cin>>a;
cout<<f[a]<<endl;
}
return 0;
}
Going up the stairs is also a Fibonacci sequence.
Recursion formula: a[i] = a[i-1]+a[i-2]+a[i-3]; Recursion boundary:
a[1]=1; a[2]=2; a[ 3]=4;
#include<iostream>
using namespace std;
long long a[100];
int main(){
a[1]=1;
a[2]=2;
a[3]=4;
int n;
do{
cin>>n;
if(n==1 || n==2 || n==3){
cout<<a[n]<<endl;
}
else{
for(int i=4;i<=n;i++){
a[i] = a[i-1]+a[i-2]+a[i-3];
}
if(n==0){
return 0;
}
cout<<a[n]<<endl;
}
}while(n!=0);
return 0;
}
1. Enter the n*n value of the two-dimensional array;
2. If the coordinates of a certain room are infected, then the upper, lower, left and right sides of the person's room will be marked as infected, and an additional judgment will be made on the empty room as not infected;
3. Traverse the room once and if it is marked as infected , then turn healthy people in the room into infected!
4. m times, that is, repeat steps 2 and 3;
5. Finally count the number of infected people on the m day and output
//1191：流感传染
#include<bits/stdc++.h>
using namespace std;
char a[105][105];
bool flag[105][105]={
0};
int main(){
int n,m,cnt=0;
cin>>n;
for(int i=1;i<=n;i++){
//输入第一天值
for(int j=1;j<=n;j++){
cin>>a[i][j];
}
}
cin>>m;
for(int k=2;k<=m;k++){
// 第2天到第m天
for(int i=1;i<=n;i++){
//遍历一遍房间
for(int j=1;j<=n;j++){
if(a[i][j]=='@'){
//房间的人是感染则
//上下左右有人的房间标记感染
if(a[i-1][j]=='.') flag[i-1][j]=true;
if(a[i+1][j]=='.') flag[i+1][j]=true;
if(a[i][j-1]=='.') flag[i][j-1]=true;
if(a[i][j+1]=='.') flag[i][j+1]=true;
}
}
}
for(int i=1;i<=n;i++){
//再遍历一遍房间，
for(int j=1;j<=n;j++){
if(flag[i][j]==true){
// 对感染的房间里的人改变状态，
a[i][j]='@'; //标记生病
}
}
}
}
//第m天后，统计感染的的人数，并输出
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(a[i][j]=='@'){
cnt++;
}
}
}
cout<<cnt;
return 0;
}
This question is very similar to "Pawn Crossing the River", and is simpler. Prerequisite: You can only go up or to the right! , that is, the rows will become larger or the columns will become larger. Calculate the sum of the routes taken. Note that there is only one way to move in a single row and single column!
- 递推式：a[i][j]=a[i-1][j] + a[i][j-1];
#include<bits/stdc++.h>
using namespace std;
int a[21][21];
int main(){
int m,n;
cin>>m>>n;
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(i==1 && j!=1){
//单行(1,1) (1,2)的路径数1
a[1][j]=1;
}
else if(j==1 && i!=1){
//单列(1,1) (2,1)的路径数1
a[i][1]=1;
}
else{
//往上走或往右走的递推公式
a[i][j]=a[i-1][j] + a[i][j-1];
}
}
}
cout<<a[m][n];
return 0;
}