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http://www.codeforces.com/gym/100827/attachments
Time Limits: 5000 MS Memory Limits: 200000 KB
64-bit interger IO format: %lld Java class name: Main
A Hill Number is a number whose digits possibly rise and then possibly fall, but never fall and then rise.
• 12321 is a hill number.
• 101 is not a hill number.
• 1111000001111 is not a hill number.
Given an integer n, if it is a hill number, print the number of hill numbers less than it. If it is not a hill number, print -1.
Input will start with a single line giving the number of test cases. Each test case will be a single positive integer on a single line, with up to 70 digits. The result will always fit into a 64-bit long.
For each test case, print -1 if the input is not a hill number. Print the number of hill numbers less than the input value if the input value is a hill number.
5 10 55 101 1000 1234321
#include<set> #include<map> #include<queue> #include<stack> #include<cmath> #include<string> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define INF 1<<30 #define MOD 1000000007 #define ll long long #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define pi acos(-1.0) using namespace std; const int MAXN = 100; ll dp[MAXN][10][5]; int digit[MAXN]; char s[MAXN]; int ok(char s[]) { int len = strlen(s); int flag = 0; for(int i = 1; i < len; i++){ if(s[i] > s[i-1] && flag){ return 0; } else if(s[i] < s[i-1] && !flag){ flag = 1; } } return 1; } ll dfs(int len,int w,int ismax,int pa) { if(len == 0)return 1; if(!ismax && dp[len][pa][w]){//第i位 为pa数字 状态为w的时候的个数 return dp[len][pa][w]; } ll ans = 0; ll fans = 0; int maxv = ismax ? digit[len] : 9; for(int i = 0; i <= maxv; i++){ if(w == 1){ if(i >= pa){ ans += dfs(len-1,1,ismax && i == maxv,i); } else { ans += dfs(len-1,3,ismax && i == maxv,i); } } else if(w == 2){ if(i > pa)continue; else { ans += dfs(len-1,2,ismax && i == maxv,i); } } else if(w == 3) { if(i > pa)continue; else { ans += dfs(len-1,3,ismax && i == maxv,i); } } } if(!ismax)dp[len][pa][w] = ans; return ans; } void solve() { if(!ok(s)){ printf("-1\n"); return ; } int len = 0; memset(dp,0,sizeof(dp)); for(int i = strlen(s) - 1; i >= 0; i--){ digit[++len] = s[i] - ‘0‘; } printf("%lld\n",dfs(len,1,1,-1) - 1); } int main() { int t; scanf("%d",&t); while(t--){ scanf("%s",s); solve(); } return 0; }
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原文地址：http://www.cnblogs.com/sweat123/p/5538657.html
