Corner Valley conjecture---memory search

2969 Corner Valley conjecture

time limit: 1 s space limit: 32000 KB

Title Description Description

The so-called Angular Valley conjecture, that is, given a positive integer n, the following two transformations are repeated on N:
1) If n is an even number, divide by 2;
2) If n is an odd number, multiply by 3 plus 1.
The final result is always 1.

The number of transformations we need to make from N to 1 is called the transformation length of N, as the transformation of the number 7 is:

7-22-11-34-17-52-26-13-40-20-10-5-16-8-4-2-1

A total of 16 transformations were made, thus 7 of the transformation length was 16.

Wish is now interested in the longest transformation length in a given interval, but the hand counts too much, so he finds you in the informatics contest, can you help him?

Enter a description input Description

Each test point contains multiple sets of data, and the first row is a number T, representing the number of data.
The second line to line t+1, two numbers a, b for each line, represents the longest transformation length of a and B (including A, b).

outputs description output Description

Output format
T row, each row outputs the longest transformation length for each interval of the input data.

sample input to sample

2
1 7
9 20

Sample output Sample outputs

16
20

data size & Hint

Data range
1 <= T <= 100
1 <= A, b <= 10^8
Interval length not exceeding 10^5

The idea is to open a large array of pre-storage results, and then the memory of the search is much faster, between the magnitude of the point

To use a 64-bit integer

#include <stdio.h> #include <string.h>int a[1000000];int dfs (long long i) {if (I < 1000000) {if (a[i]! =-1) return a[i];} if (i% 2 = = 0) return Dfs (I/2) + 1;else return DFS (i * 3 + 1) + 1;} int main () {Long Long i;memset (A,-1, sizeof (a)); A[1] = 0;for (i = 2; i < 1000000; i++) {A[i] = DFS (i);} int m, n, J, T, sum;scanf ("%d", &t), while (t--) {int tmp;scanf ("%d%d", &m, &n), if (M > N) {tmp = M;m = N;n = TMP;} sum = 0;for (i = m; i <= N; i++) {tmp = DFS (i); if (tmp > SUM) sum = tmp;} printf ("%d\n", sum);} return 0;}

Corner Valley conjecture---memory search