Codeforces Round #446 (Div. 2)

A.找出最大的两个数 注意ll

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3e5 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
ll a[100005],b[100005];
int main()
{
        //freopen("in.txt", "r", stdin);
        int n;
        ll sum=0;
        cin >> n;
        for(int i=1;i<=n;i++)
        {
        cin >> a[i];
        sum+=a[i];
        }
        for(int i=1;i<=n;i++)
        cin >> b[i];
        sort(b+1,b+n+1);
        if(b[n]+b[n-1]>=sum)
        cout<<"YES"<<endl;
        else
        cout<<"NO"<<endl;
}

B.维护left

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3e5 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
int a[1000005];
int main()
{
        //freopen("in.txt", "r", stdin);
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                cin >> a[i];
        }
        int remain = 0;
        int left = n+1;
        for (int i = n; i >= 1; i--)
        {
                if (i < left)
                {
                        remain++;
                        left = i;
                }
                left = min(left, max(1, i - a[i]));
        }
        cout << remain << endl;
}

C.找出搞出1的最少次数

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
const double EPS = 1.0e-8;
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3e5 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//next_permutation
int a[2005];
int b[2005][2005];
int anser = 0;
int n;
int gcd(int x, int y)
{
        if (y == 0)
        {
                return x;
        }
        else
        {
                return (gcd(y, x % y));
        }
}
void dfs(int x)
{
        int flag = 0;
        int now = n - x;
        for (int i = 1; i <= now; i++)
        {
                b[x][i] = gcd(b[x - 1][i], a[i + x]);
                if (b[x][i] == 1)
                {
                        anser = x;
                        flag = 1;
                        break;
                }
        }
        if (!flag)
        {
                dfs(x + 1);
        }
}
int main()
{
        //freopen("in.txt", "r", stdin);
        int flag = 0;
        int finalgcd;
        int cur=0;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                cin >> a[i];
                if (a[i] == 1)
                {
                        cur++;
                        flag = 1;
                }
        }
        if (flag)
        {
                cout << n - cur << endl;
                return 0;
        }
        flag = 0;
        for (int i = 1; i < n; i++)
        {
                b[1][i] = gcd(a[i], a[i + 1]);
                if (i == 1)
                {
                        finalgcd = b[1][1];
                }
                else
                {
                        finalgcd = gcd(finalgcd, a[i + 1]);
                }
                if (b[1][i] == 1)
                {
                        flag = 1;
                        break;
                }
        }
        //TS;
        if (flag)
        {
                cout << n << endl;
        }
        else
        {
                if (finalgcd != 1)
                {
                        cout << -1 << endl;
                }
                else
                {
                        //TS;
                        dfs(2);
                        cout << anser + n - 1 << endl;
                }
        }
}

!!!D.构造题  给定一个数组a,将a中的元素调换位置变成b数组，使得对于每一个下标子集，a中对应下标的元素和不等于b中对应下标的元素和.

正解:

把a序列做一个排序，设为c序列，然后我们开始构造b序列 

b[i]=c[(j+1)%n]，满足c[j]=a[i]

证明：

a) 假设我在a序列中取得子序列不包含最大值的话。那么一定满足a的子序列的和总小于b的子序列的和。

b) 假设我在a序列中取得子序列包含最大值的话。那么只要子序列的长度小于n，就一定会有a的子序列的和总大于b的子序列的和。

综上所述，a、b中挑选对应位置的子序列，只要长度<n，那么和就永远不可能相等

#include <bits/stdc++.h>
using namespace std;
const int maxn=30;
int a[maxn];
int b[maxn];
int n;
bool cmp(int a,int b)
{
    return a>b;
}
int fid(int x)
{
    for(int i=0;i<n;i++){
        if(b[i]==x){
            return i;
        }
    }
}
int main()
{
//    freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
        b[i]=a[i];
    }
    sort(b,b+n,cmp);
    int index;
    for(int i=0;i<n;i++){
        index=fid(a[i]);
        if(index==0){
            cout<<b[n-1]<<" ";
        }
        else{
            cout<<b[index-1]<<" ";
        }
    }
    cout<<endl;
}