1 solutions
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0
C++ :
#include<bits/stdc++.h> using namespace std; int n,a[1005],ans,at,b,t; int main() { cin>>n; for(int i=0;i<n;i++) {cin>>b;a[b]++;} while(1) { int q=0;//标记 for(int i=t;i<=1000;i++) if(a[i]>0) {a[i]--;t++;at++;q=1;break;} //个数减一,往下一个找,找到个数加一,标记 if(q==0) {ans++;t=0;}//一直找到没有为止 if(at==n) break;//找到个数是n个,退出 } cout<<ans+1;//因为最后一次q不等于0,没有累加1 return 0; }
- 1
Information
- ID
- 977
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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