【DP】 codeforces 466D Increase Sequence

很巧妙的DP。。。转化成每次都减1.。。用一维很巧妙的解决了。。。

#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 2005
#define maxm 40005
#define eps 1e-10
#define mod 1000000007
#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 qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
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();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head
int a[maxn];
int n, m;
void read(void)
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 1; i <= n; i++) a[i] = m - a[i];
}
void work(void)
{
for(int i = 1; i <= n; i++) if(a[i] < 0) {
printf("0n");
return;
}
a[0] = a[n+1] = 0;
LL ans = 1;
for(int i = 1; i <= n + 1; i++) {
if(a[i] == a[i-1] + 1) continue;
if(a[i] == a[i-1]) ans = ans * (a[i] + 1) % mod;
else if(a[i] == a[i-1] - 1) ans = ans * a[i-1] % mod;
else {
printf("0n");
return;
}
}
printf("%I64dn", ans);
}
int main(void)
{
read();
work();
return 0;
}

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