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luogu 1192
题目分析:
定义f[i]为走到第i个台阶的方案数
考虑状态转移:f[i]+=f[i-j] ,1<=j<=k;
第i号台阶可以由第i-j个台阶走过来,方案数就是它们的和
考虑边界,f[0]=1;
注意每次取模,不然会炸
Code:
#include <bits/stdc++.h>
using namespace std;
#define maxn 100010
#define maxk 110
int n,k,f[maxn];
inline void init_() {
freopen("a.txt","r",stdin);
}
inline int read_() {
int x=0,f=1;
char c=getchar();
while(c<'0'||c>'9') {
if(c=='-') f=-1;
c=getchar();
}
while(c>='0'&&c<='9') {
x=x*10+c-'0';
c=getchar();
}
return x*f;
}
inline void readda_() {
n=read_();k=read_();
}
inline void clean_() {
memset(f,0,sizeof(f));
}
inline void work_() {
clean_();
f[0]=1;
for(int i=1;i<=n;i++) {
for(int j=1;j<=k;j++) {
if(i-j<0) break;
f[i]+=f[i-j];
f[i]%=100003;
}
}
printf("%d",f[n]%100003);
}
int main() {
init_();
readda_();
work_();
return 0;
}