Problem DescriptionOne integer number x is called "Mountain Number" if:
(1) x>0 and x is an integer;
(2) Assume x=a[0]a[1]...a[len-2]a[len-1](0≤a[i]≤9, a[0] is positive). Any a[2i+1] is larger or equal to a[2i] and a[2i+2](if exists).
For example, 111, 132, 893, 7 are "Mountain Number" while 123, 10, 76889 are not "Mountain Number".
Now you are given L and R, how many "Mountain Number" can be found between L and R (inclusive) ?
InputThe first line of the input contains an integer T (T≤100), indicating the number of test cases.
Then T cases, for any case, only two integers L and R (1≤L≤R≤1,000,000,000).
Output Sample Input Sample Output#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int t; int bit[20],len,l,r; int dp[20][10][2]; int dfs(int pos,int pre,int odd,int zero,int doing)//pos-当前位置，pre前一个数字，odd，现在位置的奇偶性，zero是否还是前导0，doing是否美剧导了边界 { if(pos==-1) return 1; if(dp[pos][pre][odd]!=-1 && !doing) return dp[pos][pre][odd]; int end = doing?bit[pos]:9; int ans = 0; for(int i = 0; i<=end; i++) { if(!(i||zero)) ans+=dfs(pos-1,9,0,zero||i,doing&&i==end); else if(odd && pre<=i) ans+=dfs(pos-1,i,!odd,zero||i,doing&&i==end); else if(!odd && pre>=i) ans+=dfs(pos-1,i,!odd,zero||i,doing&&i==end); } if(!doing) dp[pos][pre][odd] = ans; return ans; } int cal(int x) { len = 0; while(x) { bit[len++] = x%10; x/=10; } return dfs(len-1,9,0,0,1);//刚刚开始在前面加个9，方便后面判断 } int main() { scanf("%d",&t); while(t--) { memset(dp,-1,sizeof(dp)); scanf("%d%d",&l,&r); printf("%d\n",cal(r)-cal(l-1)); } return 0; }
原文地址：http://blog.csdn.net/libin56842/article/details/41847295
