"尚學堂杯"哈爾濱理工大學第七屆程式設計競賽
阿新 • • 發佈：2019-02-08
| C.Collection Game | |||||
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| Description | |||||
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POI and POJ are pair of sisters, one is a master in “Kantai Collection”, another is an excellent competitor in ACM programming competition. One day, POI wants to visit POJ, and the pace that between their homes is made of square bricks. We can hypothesis that
POI’s house is located in the NO.1 brick and POJ’s house is located in the NO.n brick. For POI, there are three ways she can choose to move in every step, go ahead one or two or three bricks. But some bricks are broken that couldn’t be touched. So, how many
ways can POI arrive at POJ’s house? | |||||
| Input | |||||
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There are multiple cases. In each case, the first line contains two integers N(1<=N<=10000) and M (1<=M<=100), respectively represent the sum of bricks, and broke bricks. Then, there are M number in the next line, the K-th number a[k](2<=a[k]<=n-1) means the brick at position a[k] was broke. |
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| Output | |||||
| Please output your answer P after mod 10007 because there are too many ways. | |||||
| Sample Input | |||||
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5 1 3 |
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| Sample Output | |||||
| 3 |
#include <iostream> #include<stdio.h> #include<cstring> const int mod=10007; using namespace std; int a[10001],b[10001],c[10001]; int main() { int n,m; while(scanf("%d%d",&n,&m)==2) { memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); for(int i=1;i<=m;i++) { scanf("%d",&a[i]); b[a[i]]=1; } c[1]=1; if(!b[2]) { c[2]=1; } if(!b[3]) { c[3]=c[2]+c[1]; } if(!b[4]) { c[4]=c[3]+c[2]+c[1]; } for(int i=5;i<=n;i++) { if(!b[i]) { c[i]=(c[i-1]+c[i-2]+c[i-3])%mod; } } cout<<c[n]<<endl; } return 0; }
自己之前寫的程式碼是遞迴超時，然後這裡給了個官方題解#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[100005], vis[100005];
const int mod = 10007;
int main() {
int n, m;
while(~scanf("%d%d", &n, &m)) {
memset(vis, 0, sizeof(vis));
memset(dp, 0, sizeof(dp));
for(int i = 0, x; i < m; i++) {
scanf("%d", &x);
if(x < n) vis[x] = 1;
}
dp[1] = 1;
for(int i = 2; i <= n; i++) {
if(!vis[i]) {
dp[i] = (dp[i] + dp[i-1])%mod;
if(i > 2) dp[i] = (dp[i] + dp[i-2])%mod;
if(i > 3) dp[i] = (dp[i] + dp[i-3])%mod;
} else dp[i] = 0;
}
printf("%d\n", dp[n]);
}
return 0;
}