HDU 6148 Valley Number
作者:互联网
记录一下前面的递增递减状态,再记录一下前一个数,3维状态。
下附代码:
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const ll MOD=1e9+7; 5 ll dim[105]; 6 ll dp[105][2][10]; 7 ll dfs(int x,int up,int st,int jb,int fz){ 8 if (!x) { 9 return 1; 10 } 11 if (!jb && !fz && dp[x][up][st]!=-1 && up!=-1 && st!=-1) return dp[x][up][st]; 12 ll ret=0; 13 int maxn=9; 14 if (jb) maxn=dim[x]; 15 for (int i=0; i<=maxn; i++){ 16 if (fz){ 17 if (i==0) ret=(ret+dfs(x-1,-1,-1,(jb==1 && maxn==i),1))%MOD; 18 else ret=(ret+dfs(x-1,-1,i,(jb==1 && maxn==i),0))%MOD; 19 } 20 else { 21 if (up==1 && i<st) continue; 22 if (i>st) ret=(ret+dfs(x-1,1,i,(jb==1 && maxn==i),0))%MOD; 23 else if (i==st) ret=(ret+dfs(x-1,up,st,(jb==1 && maxn==i),0))%MOD; 24 else ret=(ret+dfs(x-1,0,i,(jb==1 && maxn==i),0))%MOD; 25 } 26 } 27 if (!jb && !fz && up!=-1 && st!=-1) return dp[x][up][st]=ret; 28 return ret; 29 } 30 int main(){ 31 memset(dp,-1,sizeof(dp)); 32 int T; 33 scanf("%d",&T); 34 while (T--){ 35 string a; 36 cin>>a; 37 for (int i=0; i<a.length(); i++) 38 dim[a.length()-i]=a[i]-'0'; 39 printf("%lld\n",(dfs(a.length(),-1,-1,1,1)-1+MOD)%MOD); 40 } 41 }View Code
标签:HDU,Valley,int,up,ret,st,&&,6148,jb 来源: https://www.cnblogs.com/i-caigou-TT/p/14372006.html