codefroces round 320 div2 题解

A： 水题

#include <bits/stdc++.h>using namespace std;#define rep(i,a,n) for (int i=a;i<=n;i++)#define per(i,n,a) for (int i=n;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const ll mod=1000000007;ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}#define Fast_IO ios_base::sync_with_stdio(0);cin.tie(0)int n;int main(){cin>>n;int tmp = 30;int res = 0;while(n){int t = 1<<tmp;if(t > n) tmp--;else n -= t,res++;}cout<<res<<endl;return 0;}

B： 至今也感觉贪心不对。。。

#include <iostream>#include <algorithm>#include <cstdio>#include <cstdlib>#include <cstring>#define N 1000005using namespace std;struct node{int x,y,v;const bool operator < (const node old) const{return v > old.v;}}t[N];bool visit[N];int n,ans[N];int main(){while(scanf("%d",&n) == 1){int i,j,k = 0;for(i = 2;i <= n * 2;i++){for(j = 1;j < i;j++){scanf("%d",&t[k].v);t[k].x = i;t[k].y = j;k++;}}int cont = n;sort(t,t + k);memset(visit,0,sizeof(visit));for(i = 0;i <= n*2*(n*2-1);i++){//cout << t[i].v << endl;if(cont == 0)break;if(visit[t[i].x] == 0 && visit[t[i].y] == 0){visit[t[i].x] = visit[t[i].y] = 1;ans[t[i].x] = t[i].y;ans[t[i].y] = t[i].x;cont--;}}for(i = 1;i <= 2 * n;i++){if(i == 1)cout << ans[i];else cout << " " << ans[i];}putchar('\n');}return 0;} 

C：推公式。或者二分

#include <bits/stdc++.h>using namespace std;#define rep(i,a,n) for (int i=a;i<=n;i++)#define per(i,n,a) for (int i=n;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const ll mod=1000000007;ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}#define Fast_IO ios_base::sync_with_stdio(0);cin.tie(0)int n;int a,b;int main(){cin>>a>>b;if(a < b) {cout<<-1<<endl; return 0;} int t = ((a/b+1)&1)?a/b:a/b+1;double ans = (a+b)*(1.0)/t;printf("%.9f\n", ans);return 0;}

D： 预处理前缀后缀乱搞

#include <bits/stdc++.h>using namespace std;#define rep(i,a,n) for (int i=a;i<=n;i++)#define per(i,n,a) for (int i=n;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;ll powmod(ll a,ll b) {ll res=1;for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}#define Fast_IO ios_base::sync_with_stdio(0);cin.tie(0)#define EPS 1e-8#define LL long long#define MAXN 200010LL l[MAXN] = {0}, r[MAXN] = {0};LL a[MAXN];int main () {    int n, k, x;    scanf("%d%d%d", &n, &k, &x);    for(int i = 1; i <= n; i++)        scanf("%I64d", &a[i]);    for(int i = 1; i <= n; i++) {        l[i] = l[i - 1] | a[i];    }    for(int i = n; i >= 1; i--) {        r[i] = r[i + 1] | a[i];    }    LL maxv = 0;    for(int i = 1; i <= n; i++) {        LL xx = l[i - 1] | r[i + 1];        LL now = a[i];        for(int j = 0; j < k; j++){            now *= x;        }        maxv = max(maxv, now | xx);    }    cout << maxv << endl;    return 0;}

E： 三分+最大字段和

#include <bits/stdc++.h>using namespace std;#define rep(i,a,n) for (int i=a;i<=n;i++)#define per(i,n,a) for (int i=n;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define SZ(x) ((int)(x).size())#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;ll powmod(ll a,ll b) {ll res=1;for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}#define Fast_IO ios_base::sync_with_stdio(0);cin.tie(0)int n;double a[200010];#define EPS 1e-8const long long INF = 0x3f3f3f3f;double f[200010];double solve(double x){    double ans1=0;    f[1]=a[1]-x;    for(int i=2;i<=n;i++)    {        if(f[i-1]<0)            f[i]=a[i]-x;        else            f[i]=f[i-1]+a[i]-x;        ans1=max(ans1,f[i]);    }    f[1]=(a[1]-x);    ans1=max(ans1,-f[1]);    for(int i=2;i<=n;i++)    {        if(f[i-1]>0)            f[i]=(a[i]-x);        else            f[i]=f[i-1]+(a[i]-x);        ans1=max(ans1,-f[i]);    }    return ans1;}int main(){cin>>n;rep(i, 1, n){scanf("%lf", &a[i]);}double l = -10000*(1.0);double r = 10000*1.0;int cnt = 300;double ans;while(cnt--){double mid = (2*l+r)/3;double mmid = (l+2*r)/3;double res1 = solve(mid);double res2 = solve(mmid);if(res1 >= res2){l = mid;ans = mid;}else{ans = mmid;r = mmid;}}double res = 0;double ret = 0;for(int i=1; i<=n; i++){res += a[i] - ans;if(res - EPS < 0)res = 0; ret = max(ret, res);}printf("%.7f\n", ret);return 0;}