ACWing 1307. Stall and Bull (combination number, DP, prefix sum)

1307. Stall and Cow

Idea One

• $f\left[i\right]$ means there is$The$ number of plans for$i$ cows and the last cow must be a bull.
• $f[i]={\sum }_0^{i-k-1}f[i]$ $(s[i-k,i-1\left]\right)$ Only cows are allowed.
• The final answer: ${\sum }_0^{n}f\left[i\right]$ , do things in categories, the principle of addition.

#include <stdio.h>
using namespace std;

const int mod = 5000011,size = 1e5+10;
int n,k,f[size] = {
    
    1},s[size] = {
    
    1};

int main(){
    
    
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
    
    
        f[i] = i-k-1>=0?s[i-k-1]:f[0];
        s[i] = (s[i-1]+f[i])%mod;
    }
    printf("%d\n",s[n]);
    return 0;
}

Idea two

• Use f [i] f[i] directly$f\left[i\right]$ means there isiiNumber of plans for$i$ cattle
  • If $i$ cow is a cow, then$i-1$ head can be placed casually according to the requirements of the subject.
  • If $i$ cow is a bull,$s[i-k]$~$s[i-1]$ None can be bulls, only cows.
    So at this time$i-k-1$ cow is placed randomly according to the requirements of the subject.
    $f[i]=f[i-1]+f[i-k-1]$
• The answer is $f[n]$

#include <stdio.h>
using namespace std;

const int mod = 5000011,size = 1e5+10;
int n,k,f[size] = {
    
    1};

int main(){
    
    
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
    
    
        f[i] = f[i-1];
        if(i-k-1>=0) f[i] = (f[i]+f[i-k-1])%mod;
        else f[i] = (f[i]+1)%mod;
    }
    printf("%d\n",f[n]);
    return 0;
}