Codeforces 950 010子序列拆分 数列跳棋

A

B

a,b两个序列分成K段 每一段的值相等

#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; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int  maxn = 100005;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll mod = 3e7;
ll num[maxn];
ll num2[maxn];
//priority_queue<int, vector<int>, less<int>> que;
int main()
{
        int n, m;
        ll now;
        cin >> n >> m;
        for (int i = 1; i <= n; i++)
        {
                scanf("%lld", num+i);
                //pre[i] = pre[i - 1] + now;
        }
        for (int i = 1; i <= m; i++)
        {
                scanf("%lld", num2+i);
                //pre2[i] = pre2[i - 1] + now;
        }
        int cur = 1;
        int cur2 = 1;
        ll sum1 = num[1];
        ll sum2 = num2[1];
        int anser = 0;
        while (cur != n + 1)
        {
                if (sum1 == sum2)
                {
                        anser++;
                        sum1+=num[++cur];
                        sum2+=num2[++cur2];
                        continue;
                }
                if(sum1<sum2)
                {
                        sum1+=num[++cur];
                }
                else
                {
                        sum2+=num2[++cur2];
                }
        }
        cout<<anser<<endl;
}

C

问子序列可否全部分成0或者010 01010这种01相间的序列

用Vector和一个int指针维护 int之前的vector表示以0结尾的 之后的表示以1结尾的

如果指针在插入一个1前是在0或者全部插入完还有以1结尾的就return false

#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; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int  maxn = 100005;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll mod = 3e7;
vector<int> ans[200005];
int anser = 0;
int cnt = 0;
//priority_queue<int, vector<int>, less<int>> que;
int main()
{
        string a;
        cin >> a;
        int len = a.size();
        for (int i = 0; i < len; i++)
        {
                if (a[i] == '0')
                {
                        ans[++cnt].pb(i + 1);
                }
                else
                {
                        if (cnt <= 0) //!!!!!!!!!!!!
                        {
                                cout << -1 << endl;
                                return 0;
                        }
                        ans[cnt--].pb(i + 1);
                }
                anser = max(anser, cnt);
        }
        if (cnt != anser)
        {
                cout << -1 << endl;
                return 0;
        }
        cout << anser << endl;
        for (int i = 1; i <= anser; i++)
        {
                int lenth = ans[i].size();
                cout << lenth << " ";
                for (int j = 0; j < lenth; j++)
                {
                        cout << ans[i][j] << " ";
                }
                cout << endl;
        }
}

D

思维题 可以找规律做 也可以推公式做

很容易想到原本序列前N/2个数的位置是不会变的 因为这些数一旦跳了 最后序列的长度就会小于N  这些数在最后的位置下标为奇数位

然后其他跳的数 假设他跳之后的位置为P 则现在在他左边的P/2个数是原数组原来就在他左边的  

右边有N-P/2个数（包括他自己）是连续的   也就是说他之前的位置往前进了N-P/2个块

所以公式就是P=P/2+N  一直循环直到P为奇数为止 这就是最终序列的数对应原来序列的位置

最后答案就是P/2+1

#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; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
priority_queue<int, vector<int>, less<int>> que;
ll mod = 3e7;
const int MAXN = 505;
int main()
{
    ll n;
    ll q;
    cin >> n >> q;
    while(q--)
    {
        ll now;
        cin >> now;
        while(!(now%2))
        {
            now=now/2+n;
        }
        cout<<now/2+1<<endl;
    }
}