bzoj 1026: [SCOI2009]windy数
D（不）P（会）类
1 #include<bits/stdc++.h> 2 #define N 200005 3 #define LL long long 4 #define inf 0x3f3f3f3f 5 #define ls tr[x][0] 6 #define rs tr[x][1] 7 using namespace std; 8 inline int ra() 9 { 10 int x=0,f=1; char ch=getchar(); 11 while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();} 12 while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();} 13 return x*f; 14 } 15 int f[12][10]; 16 int dp[12]; 17 void init() 18 { 19 memset(f,0,sizeof(f)); 20 for (int i=0; i<10; i++) f[1][i]=1; 21 for (int i=2; i<=10; i++) 22 { 23 for (int j=0; j<10; j++) 24 { 25 for (int k=0; k<10; k++) 26 { 27 if (abs(j-k)>=2) 28 f[i][j]+=f[i-1][k]; 29 } 30 } 31 } 32 } 33 int len; 34 int solve(int a) 35 { 36 len=0; 37 while (a) 38 { 39 dp[len++]=a%10; 40 a/=10; 41 } 42 int ans=0; 43 for (int i=1; i<len; i++) 44 for (int j=1; j<10; j++) 45 ans+=f[i][j]; 46 for (int i=1; i<dp[len-1]; i++) 47 ans+=f[len][i]; 48 for (int i=len-1; i>=1; i--) 49 { 50 for (int j=0; j<dp[i-1]; j++) 51 if (abs(j-dp[i])>=2) 52 ans+=f[i][j]; 53 if (abs(dp[i]-dp[i-1])<2) break; 54 } 55 return ans; 56 } 57 int main() 58 { 59 init(); int a,b; 60 while (~scanf("%d%d",&a,&b)) 61 { 62 cout<<solve(b+1)-solve(a); 63 } 64 return 0; 65 }
