Vasily has a number a, which he wants to turn into a number b. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number x by 2·x);
- append the digit 1 to the right of current number (that is, replace the number x by 10·x + 1).
You need to help Vasily to transform the number a into the number b using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform a into b.
The first line contains two positive integers a and b (1 ≤ a < b ≤ 109) — the number which Vasily has and the number he wants to have.
If there is no way to get b from a, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer k — the length of the transformation sequence. On the third line print the sequence of transformations x1, x2, ..., xk, where:
- x1 should be equal to a,
- xk should be equal to b,
- xi should be obtained from xi - 1 using any of two described operations (1 < i ≤ k).
If there are multiple answers, print any of them.
2 162
YES 5 2 4 8 81 162
4 42
NO
100 40021
YES
5
100 200 2001 4002 40021
这题的坑点只是太隐蔽了，n虽然给了10^9，让你以为int内可以处理到，但是*10+1后就可能会爆int，一直以为是数组开小的导致的M，结果是数据导致的。。
bfs
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> using namespace std; const int N = 100; struct node { long long num[N], x; int step;; }; int main() { long long a, b; while(scanf("%I64d %I64d", &a, &b)!=EOF) { queue<node>q; while(!q.empty()) { q.pop(); } node e; e.step=1, e.x=a, e.num[1]=a; q.push(e); int flag=0; while(!q.empty()) { node u; u = q.front(); q.pop(); if(u.x==b) { flag=1; printf("YES\n"); printf("%d\n",u.step); for(int i=1;i<=u.step;i++) { printf("%I64d%c",u.num[i],i==u.step?'\n':' '); } break; } if(u.x>b) continue; node u2 = u; u2.x = u.x*10+1; u2.step = u.step+1; u2.num[u2.step]=u2.x; q.push(u2); u.x = u.x*2; u.step = u.step+1; u.num[u.step]=u.x; q.push(u); } if(flag==0) printf("NO\n"); } return 0; }
dfs #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> using namespace std; const int N = 1000; long long num[N]; int k; int a, b, flag; int dfs( long long x) { if(flag==1) return 0; if(x==b) { flag=1; num[k++]=b; return 1; } if(x>b) return 0; if(dfs(x*10+1)) { num[k++]=x; return 1; } if(dfs(x*2)) { num[k++]=x; return 1; } return 0; } int main() { while(scanf("%d %d", &a, &b)!=EOF) { flag=0; k=0; dfs(a); if(flag==0) printf("NO\n"); else { printf("YES\n"); printf("%d\n",k); for(int i=k-1;i>=0;i--) { printf("%lld%c",num[i],i==0?'\n':' '); } } } return 0; }