考虑：
因此偶数可以任选，但是奇数必须保证偶数个，因此先选出所有偶数接着对奇数排序，选尽量大的偶数个即可
#include <set>#include <map>#include <stack>#include <queue>#include <math.h>#include <cstdio>#include <string>#include <bitset>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>#include <unordered_map>using namespace std;#define fi first#define se second#define pb push_back#define ins insert#define lowbit(x) (x&(-x))#define mkp(x,y) make_pair(x,y)#define mem(a,x) memset(a,x,sizeof a);typedef long long ll;typedef long double ld;typedef unsigned long long ull;typedef pair<int,int> P;const double eps=1e-8;const double pi=acos(-1.0);const int inf=0x3f3f3f3f;const ll INF=1e18;const int Mod=1e9 7;const int maxn=2e5 10;int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int n; cin>>n; ll sum=0; vector<int> res; for(int i=1,x;i<=n;i ){ cin>>x; if(x&1) res.push_back(x); else sum =x; } sort(res.begin(),res.end()); int k=res.size(); if(k&1){ for(int i=1;i<k;i ) sum =res[i]; }else{ for(int i=0;i<k;i ) sum =res[i]; } cout<<sum<<endl; return 0;}因为有偶数个点，而且上下两个圆平面中心连线一定为某坐标轴，因此该中线将点的该轴坐标分为两部分，那么我们枚举三种坐标找到该轴
最后只需求出一个正多边形上某点到其他点最大距离为上下底面半径（下面好像有点问题，我也看不太懂了）
#include <set>#include <map>#include <stack>#include <queue>#include <math.h>#include <cstdio>#include <string>#include <bitset>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>#include <unordered_map>using namespace std;#define fi first#define se second#define pb push_back#define ins insert#define lowbit(x) (x&(-x))#define mkp(x,y) make_pair(x,y)#define mem(a,x) memset(a,x,sizeof a);typedef long long ll;typedef long double ld;typedef unsigned long long ull;typedef pair<int,int> P;const double eps=1e-8;const double pi=acos(-1.0);const int inf=0x3f3f3f3f;const ll INF=1e18;const int Mod=1e9 7;const int maxn=2e5 10;int t,n;double h1,h2;double x[105],y[105],z[105];inline int dcmp(double d){ if(fabs(d)<eps) return 0; return d>0?1:-1;}bool check(double *d){ for(int i=1;i<=2*n;i ){ int cnt=0; for(int j=1;j<=2*n;j ) if(dcmp(d[j]-d[i])==0) cnt ; if(cnt==n){ h1=d[i]; for(int j=1;j<=2*n;j ) if(d[j]!=d[i]){ h2=d[j]; break; } return 1; } } return 0;}double dis(double x1,double y1,double x2,double y2){ return sqrt((x1-x2)*(x1-x2) (y1-y2)*(y1-y2));}int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); //ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i=1;i<=2*n;i ) scanf("%lf%lf%lf",&x[i],&y[i],&z[i]); if(check(x)){ double h=fabs(h1-h2); double r=0.0; for(int i=1;i<=2*n;i ) r=max(r,dis(y[1],z[1],y[i],z[i])); r/=2.0; double v=pi*r*r*h; printf("%.6f\n",v); }else if(check(y)){ double h=fabs(h1-h2); double r=0.0; for(int i=1;i<=2*n;i ) r=max(r,dis(x[1],z[1],x[i],z[i])); r/=2.0; double v=pi*r*r*h; printf("%.6f\n",v); }else if(check(z)){ double h=fabs(h1-h2); double r=0.0; for(int i=1;i<=2*n;i ) r=max(r,dis(x[1],y[1],x[i],y[i])); r/=2.0; double v=pi*r*r*h; printf("%.6f\n",v); } } return 0;}经典题，一般是用栈来做的，但是这个字符串必须保证顺序，那么我们用栈的思想，维护n个队列也行
#include <set>#include <map>#include <stack>#include <queue>#include <math.h>#include <cstdio>#include <string>#include <bitset>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>#include <unordered_map>using namespace std;#define fi first#define se second#define pb push_back#define ins insert#define lowbit(x) (x&(-x))#define mkp(x,y) make_pair(x,y)#define mem(a,x) memset(a,x,sizeof a);typedef long long ll;typedef long double ld;typedef unsigned long long ull;typedef pair<int,int> P;const double eps=1e-8;const double pi=acos(-1.0);const int inf=0x3f3f3f3f;const ll INF=1e18;const int Mod=1e9 7;const int maxn=2e5 10;queue<string> s[205];int a[maxn];int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); string str; while(cin>>str){ int x=0; for(int i=0;i<str.size();i ){ if(str[i]>='0' && str[i]<='9'){ a[ x]=str[i]-'0'; i ; }else if(isalpha(str[i])){ string res=""; res =str[i]; s[x].push(res); }else if(str[i]==']'){ string res=""; while(s[x].size()){ res =s[x].front(); s[x].pop(); } string ans=""; for(int j=1;j<=a[x];j ) ans =res; x--; s[x].push(ans); } } string ans=""; while(s[0].size()){ ans =s[0].front(); s[0].pop(); } cout<<ans<<endl; } return 0;}联系客服
