FZU 2019 Mountain Number

Problem 2109 Mountain Number

Accept: 49    Submit: 120

Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

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 

比如：7068 ，存于a数组a={8,6,0,7}，那么第3位数就是7。

分析：dp[i][j][k]表示对于第i位，那么表示访问到第i位，此数还剩下i位。j表示第i-1位（即数字的下一位）填的数字，k=0表示当前为偶数位，否则为奇数位。

下面总结一下数位dp的通用写法：

关键在于写dfs，注意状态转移：

int dfs(int i,(其他参数),bool e){//i表示当前遍历第i位,e表示是否卡上界
	if(i==-1)return (...);//遍历结束
	if(!e&&dp[i][][]!=-1)return dp[i][][];//如果在不卡上界的情况下,dp[i][][]计算过,则直接返回,记忆化搜索
	int res=0,u=e?a[i]:9,d;//确定上界
	for(d=0;d<=u;d++){
		res+=dfs(i-1,(状态转移后的参数),e&&d==u);
	}
	return e?res:dp[i][][]=res;
}

至于cal(int n){}就不用说明了

主函数写法：

int main(){
	int T,l,r;
	scanf("%d",&T);
	memset(dp,-1,sizeof(dp));
	while(T--){
		scanf("%d%d",&l,&r);
		printf("%d\n",cal(r)-cal(l-1));
	}
	return 0;
}

本题题意：询问区间[l,r]内“Mountain Number”数有多少个。Mountain Number：指所有奇数位数字要大于等于和它相邻偶数位的数字

最后贴上本题代码：

注意本题中前导0的处理，当遍历到当前位置仍是前导0，则应把pre设为9

#include<cstdio>
#include<cstring>
int dp[12][10][2],a[12];
int dfs(int i,int pre,int parity,bool J,bool e){
	if(i==-1)return 1;
	if(!e&&dp[i][pre][parity]!=-1)return dp[i][pre][parity];
	int res=0,u=e?a[i]:9,d;
	for(d=0;d<=u;d++){
		if(!(J||d))res+=dfs(i-1,9,0,0,e&&d==u);
		else if(parity&&pre<=d)res+=dfs(i-1,d,!parity,J||d,e&&d==u);
		else if(!parity&&pre>=d)res+=dfs(i-1,d,!parity,J||d,e&&d==u);
	}
	return e?res:dp[i][pre][parity]=res;
}
int cal(int n){
	int i=0;
	while(n){a[i++]=n%10,n/=10;}
	return dfs(i-1,9,0,0,1);
}
int main(){
	int T,l,r;
	scanf("%d",&T);
	memset(dp,-1,sizeof(dp));
	while(T--){
		scanf("%d%d",&l,&r);
		printf("%d\n",cal(r)-cal(l-1));
	}
	return 0;
}

求区间[1,n]中符合条件(数字中含0)的数的个数，转态转移注意前导0；

dp[i][j][k] 遍历到i位,j取0或1,表示未出现过0和已出现过0,k取0或1,表示非前导0转态和仍为前导0转态

#include <cstdio>
#include <cstring>
#define ll long long
ll dp[20][2][2];
int a[20];
int new_s(int s,int d,bool lead){
	return !lead&&(s||d==0);
}
ll dfs(int pos,int s,bool lead,bool e){
	if(pos==-1)	return s;
	if(!e&&dp[pos][s][lead]!=-1) return dp[pos][s][lead];
	ll res=0;
	int u=e?a[pos]:9,i;
	for(i=0;i<=u;i++){
		res+=dfs(pos-1,new_s(s,i,lead),lead&&i==0,e&&i==u);
	}
	return e?res:dp[pos][s][lead]=res;
}
 
ll cal(ll n){
	int i=0;
	while(n)a[i++]=n%10,n/=10;
	return dfs(i-1,0,true,true);
}
int main(){
	ll n;
	memset(dp,-1,sizeof(dp));
	while(~scanf("%lld",&n)){
		printf("%lld\n",cal(n));
	}
	return 0;
}

备注一下另一种解法：

#include <cstdio>
#define ll long long
ll cal(ll n,bool & f){
	if(n<10)return 0;
	ll r=cal(n/10,f);
	if(n%10==0)f=true;
	return 9*r+n/10+f*(n%10-1);
}
int main() {
	ll n;
	while(~scanf("%lld",&n)){
		bool f=false;
		printf("%lld\n",cal(n+1,f));
	}
	return 0;
}

//求[1,n)中不含数字0的数
ll cal(ll n,bool & f){
	if(n<10)return n-1;
	ll r=cal(n/10,f);
	if(n%10==0)f=true;
	return 9*(r+1)+!f*(n%10-1);//考虑还有一种情况:前面都是0,所以是9*(r+1)
}