文章目录
HPU省赛训练（二）
H - Mountain Number
题目描述
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.
思路
应该是数位dp以前学习的不扎实，这次好好的补一下， 三维dp,dp[i][j][k] 代表这个数的第i位前面一位为j的时候k代表当前为是否为偶数位，如果k=1代表是偶数位，否则不是，
code
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int t ;
int v[20] ,cnt ;
int dp[20][20][2] ;
int dfs(int pos ,int per ,bool flag ,int odd) {
if ( cnt + 1 == pos ) return 1 ;
if ( !flag && dp[pos][per][odd] ) return dp[pos][per][odd] ;
int r = flag ? v[pos] : 9 ;
int ret = 0 ;
for (int i = 0 ; i <= r ; i ++ ) {
if ( odd && i >= per ) ret += dfs(pos + 1 ,i ,flag&&(i == v[pos]) ,odd^1) ;
else if ( !odd && i <= per ) ret += dfs(pos + 1 ,i ,flag&&(i == v[pos] ),odd^1 ) ;
}
if ( !flag ) dp[pos][per][odd] = ret ;
return ret ;
}
int slove(int x) {
cnt = 0 ;
while( x ) {
v[++cnt] = x % 10 ;
x /= 10 ;
}
if ( cnt == 1 ) return v[1] ;
reverse(v + 1 ,v + cnt + 1 ) ;
memset(dp ,0 ,sizeof(dp)) ;
return dfs(1 ,9 ,true ,0 ) ;
}
int main() {
cin >> t ;
while ( t -- ) {
int l ,r ;
scanf("%d %d",&l ,&r ) ;
printf("%d
",slove(r) - slove(l-1)) ;
}
return 0;
}
D - Digits Count
题目描述
Given N integers A={A[0],A[1],…,A[N-1]}. Here we have some operations:
Operation 1: AND opn L R
Here opn, L and R are integers.
For L≤i≤R, we do A[i]=A[i] AND opn (here “AND” is bitwise operation).
Operation 2: OR opn L R
Here opn, L and R are integers.
For L≤i≤R, we do A[i]=A[i] OR opn (here “OR” is bitwise operation).
Operation 3: XOR opn L R
Here opn, L and R are integers.
For L≤i≤R, we do A[i]=A[i] XOR opn (here “XOR” is bitwise operation).
Operation 4: SUM L R
We want to know the result of A[L]+A[L+1]+…+A[R].
Now can you solve this easy problem?
input
The first line of the input contains an integer T, indicating the number of test cases. (T≤100)
Then T cases, for any case, the first line has two integers n and m (1≤n≤1,000,000, 1≤m≤100,000), indicating the number of elements in A and the number of operations.
Then one line follows n integers A[0], A[1], …, A[n-1] (0≤A[i]<16,0≤i<n).
Then m lines, each line must be one of the 4 operations above. (0≤opn≤15)
output
For each test case and for each “SUM” operation, please output the result with a single line.