【杂题】 HDOJ 4972 A simple dynamic programming problem

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注意到只有连续的1,2和2,1才会改变最终结果。。注意还有很多情况。。

</pre><pre name="code" class="cpp">#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 40005#define eps 1e-10#define mod 998244353#define INF 999999999#define lowbit(x) (x&(-x))#define mp mark_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R  typedef long long LL;//typedef int LL;using namespace std;LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}// headint num[maxn];int n;void read(void){scanf("%d", &n);num[0] = 0;for(int i = 1; i <= n; i++) scanf("%d", &num[i]);}void work(int __){int ans = 1;int ok = 1;for(int i = 1; i <= n; i++) {if(abs(num[i] - num[i-1]) > 3) ok = 0;if(num[i] > 1 && num[i] == num[i-1]) ok = 0;if(!ok) break;}if(!ok) {ans = 0;printf("Case #%d: %d\n", __, ans);return;}for(int i = 1; i <= n; i++) {if(num[i-1] == 2 && num[i] == 1) ans++;if(num[i-1] == 1 && num[i] == 2) ans++;}if(num[n]) ans *= 2;printf("Case #%d: %d\n", __, ans);}int main(void){int _, __;while(scanf("%d", &_)!=EOF) {__ = 0;while(_--) {read();work(++__);}}return 0;}

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