FZU 2109 奇偶
2017-07-24 22:24
148 查看
传送门
One 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) ?
Input
The 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
For each test case, output the number of "Mountain Number" between L and R in a single line.
Sample Input
Sample Output
要求出区间内偶数为大于奇数位的数字的个数dp[i][j][k] 表示在第i个位置时,前面是j,现在这位是奇数位还是偶数位的数目
One 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) ?
Input
The 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
For each test case, output the number of "Mountain Number" between L and R in a single line.
Sample Input
3 1 10 1 100 1 1000
Sample Output
9 54 384
要求出区间内偶数为大于奇数位的数字的个数dp[i][j][k] 表示在第i个位置时,前面是j,现在这位是奇数位还是偶数位的数目
#include<iostream> #include<cstdio> #include<cstring> #include<string> using namespace std; #define me(x,y) memset(x,y,sizeof(x)); #define LL long long typedef long long ll; int a[25]; int dp[25][15][4]; int dfs(int pos,int pre,int sta,int lead,int limit)//pos 数位 pre 上一个数 //sta 奇偶 lead前导0 { int mod_x,sta_x; if(pos<=0) return 1; // if(sta<0) return 0; if(!limit && dp[pos][pre][sta]!=-1) return dp[pos][pre][sta]; int up=limit ? a[pos] : 9; int ans=0; for(int i=0;i<=up;i++) { if(!(i||lead)) ans+=dfs(pos-1,9,0,lead||i,limit&&i==up); else if(sta&&pre<=i) ans+=dfs(pos-1,i,!sta,lead||i,limit&&i==up); else if(!sta&&pre>=i) ans+=dfs(pos-1,i,!sta,lead||i,limit&&i==up); } if(!limit) dp[pos][pre][sta]=ans; return ans; } int solve(int x) { int pos=0;// while(x) { a[++pos]=x%10; x/=10; } return dfs(pos,9,0,0,1);//在最前弄一个9 } int main() { int t; int l,r; cin>>t; while(t--) { cin>>l>>r; //while(~scanf("%d%d",&l,&r)&&(l+r)) //while(~scanf("%d",&r)) me(dp,-1); printf("%d\n",solve(r)-solve(l-1)); } return 0; }