如果将1和3都放到正确的位置,2自然也在正确的位置。那么统计1,2,3的数量num1,num2,num3。再看前num1个数有几个(设x个)不是1,那么x个1肯定要移。设前num1个数有y个3,最后num3个数有z个1,那么这个过程中,最多能将min(y,z)个3移到正确位置。剩下num3-min(y,z)个3也必须移到最后num3个位置
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <iomanip> #include <cstring> #include <map> #include <queue> #include <set> #include <cassert> #include <stack> #define mkp make_pair using namespace std; const double EPS=1e-8; typedef long long lon; const lon SZ=1060,INF=0x7FFFFFFF; int arr[SZ]; int main() { std::ios::sync_with_stdio(0); //freopen("d:\\1.txt","r",stdin); lon casenum; //cin>>casenum; //for(lon time=1;time<=casenum;++time) { int n; cin>>n; for(int i=1;i<=n;++i)cin>>arr[i]; int num1=count(arr+1,arr+1+n,1); int num2=count(arr+1,arr+1+n,2); int num3=count(arr+1,arr+1+n,3); int x=0,y=0,z=0,w=0; for(int i=1;i<=num1;++i) { if(arr[i]!=1)++x; if(arr[i]==3)++z; } for(int i=num1+num2+1;i<=n;++i) { if(arr[i]!=3)++y; if(arr[i]==1)++w; } //cout<<num1<<" "<<num2<<" "<<num3<<endl; //cout<<x<<" "<<y<<" "<<z<<endl; cout<<(x+y-min(z,w))<<endl; } return 0; }