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| Time Limit: 4000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
Description
Edward works as an engineer for Non-trivial Elevators: Engineering, Research and Construction (NEERC). His new task is to design a brand new elevator for a skyscraper with h floors.
Edward has an idée fixe: he thinks that four buttons are enough to control the movement of the elevator. His last proposal suggests the following four buttons:
Initially, the elevator is on the first floor. A passenger uses the first three buttons to reach the floor she needs. If a passenger tries to move a, b or cfloors up and there is no such floor (she attempts to move higher than the h-th floor), the elevator doesn’t move.
To prove his plan worthy, Edward wants to know how many floors are actually accessible from the first floor via his elevator. Help him calculate this number.
Input
The first line of the input file contains one integer h — the height of the skyscraper (1 ≤ h ≤ 1018).
The second line contains three integers a, b and c — the parameters of the buttons (1 ≤ a, b, c ≤ 100 000 ).
Output
Output one integer number — the number of floors that are reachable from the first floor.
Sample Input
15
4 7 9
Sample Output
9
Source
1 /*by SilverN*/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #include<queue> 8 #define LL long long 9 using namespace std; 10 const long long inf = 1000000000ll * 1000000000ll + 10ll; 11 const int mxn=100010; 12 int a[5]; 13 LL h; 14 LL ans; 15 // 16 LL dis[mxn]; 17 bool inq[mxn]; 18 queue<int>q; 19 void SPFA(){ 20 for(int i=0;i<a[1];i++)dis[i]=inf; 21 q.push(1%a[1]); 22 dis[1%a[1]]=1; 23 inq[1%a[1]]=1; 24 while(!q.empty()){ 25 int now=q.front();q.pop(); 26 27 for(int i=2;i<=3;i++){ 28 int v=(now+a[i])%a[1]; 29 if(dis[v]>dis[now]+a[i]){ 30 dis[v]=dis[now]+a[i]; 31 if(!inq[v]){ 32 inq[v]=1; 33 q.push(v); 34 } 35 } 36 } 37 inq[now]=0; 38 } 39 } 40 int main(){ 41 scanf("%lld%d%d%d",&h,&a[1],&a[2],&a[3]); 42 sort(a+1,a+3+1); 43 SPFA(); 44 for(int i=0;i<a[1];i++){ 45 if(dis[i]<=h)ans+=(h-dis[i])/a[1]+1; 46 } 47 printf("%lld\n",ans); 48 return 0; 49 }
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原文地址：http://www.cnblogs.com/SilverNebula/p/5883599.html
