Codeforces Round #698 (Div. 2) A-E解题报告与解法证明_nezzar and board

Codeforces Round #698 (Div. 2) A-E解题报告与解法证明

题目解法总体概括

A Nezzar and Colorful Balls

#include <bits/stdc++.h>
using namespace std;

const int N = 110;
int a[N], f[N];

int main()
{
    int t;  cin >> t;
    while (t -- )
    {
        static int n;
        scanf("%d", &n);
        for (int i = 1; i <= n; i ++ )  scanf("%d", &a[i]);
        int res = 1;
        for (int i = 1; i <= n; i ++ )
        {
            f[i] = 1;
            for (int j = 1; j < i; j ++ )
            {
                if (a[j] >= a[i])    f[i] = max(f[i], f[j] + 1);
            }
            res = max(res, f[i]);
        }
        cout << res << endl;
    }

    return 0;
}

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---

B Nezzar and Lucky Number

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
LL x, n, d;

bool Check()
{
    if (x <= 9LL)   return x % d == 0;
    else if (x < 10 * d)
    {
        for (int i = 0; i < x; i += 10)
        {
            if ((x - i) % d == 0)
                return true;
        }
        return false;
    }
    else
        return true;
}
int main()
{
    int t;  cin >> t;
    while (t -- )
    {
        scanf("%lld%lld", &n, &d);
        for (int i = 1; i <= n; i ++ )
        {
            scanf("%lld", &x);
            if (Check())
                puts("YES");
            else
                puts("NO");
        }
    }

    return 0;
}

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---

C Nezzar and Symmetric Array

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 200010, INF = 0x3f3f3f3f3f3f3f3f3f;
LL a[N], d[N];
int n;

bool cmp(const LL &t1, const LL &t2)
{
    return t1 > t2;
}

int main()
{
    int t;  cin >> t;
    while (t -- )
    {
        scanf("%d", &n);
        for (int i = 1; i <= 2 * n; i ++ )  scanf("%lld", &d[i]);
        sort(d + 1, d + 1 + 2 * n, cmp);
        bool flag = true;
        d[2 * n + 1] = -INF;
        for (int i = 1; i <= n; i ++ )
        {
            if (d[2 * i] != d[2 * i - 1])
            {
                flag = false;   break;
            }
            if (d[2 * i] <= d[2 * i + 1])
            {
                flag = false;   break;
            }
        }
/*
        for (int i = 1; i <= 2 *  n; i ++ )
            printf("%lld, ", d[i]);
        cout << endl;
*/
        // 不满足 d 的基本性质
        if (!flag)
        {
            puts("NO");
            continue;
        }

        LL sum = 0; // 2(a1 + a2 + ... + a(i-1) )
        LL tmp;
        for (int i = 1; i <= n; i ++ )
        {
            tmp = d[2 * i] - sum;
            if (tmp % (2 * (n - i + 1)) != 0)
            {
                flag = false;
                break;
            }
            else
            {
                a[i] = tmp / (2 * (n - i + 1));
                if (a[i] <= 0)  // 必须是正数
                {
                    flag = false;   break;
                }
                sum += 2 * a[i];
            }
        }
/*
        for (int i = 1; i <= n; i ++ )
            printf("%lld, ", a[i]);
        cout << endl;
*/
        if (flag)
            puts("YES");
        else
            puts("NO");
    }


    return 0;
}

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---

D Nezzar and Board

两个数字x, y的情况

三个数字的x,y,z的情况
使用x,y两种情况的结论（引理）

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 200010;
LL a[N];
LL n, k;

LL gcd(LL a, LL b)
{
    if (b == 0) return a;
    else    return gcd(b, a % b);
}

bool cmp(const LL &t1, const LL &t2)
{
    return t1 > t2;
}

int main()
{
    int t;  cin >> t;

    while (t -- )
    {
        scanf("%lld%lld", &n, &k);

        for (int i = 1; i <= n; i ++ )
        {
            scanf("%lld", &a[i]);
        }

        sort(a + 1, a + 1 + n, cmp);

        LL d = a[1] - a[2];
        for (int i = 3; i <= n; i ++ )
        {
            d = gcd(d, a[1] - a[i]);
        }

        if ((k - a[1]) % d == 0) puts("YES");
        else    puts("NO");
    }

    return 0;
}

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---

E Nezzar and Binary String

#include <bits/stdc++.h>
#define lson (pos << 1)
#define rson ((pos << 1) + 1)
using namespace std;

const int N = 200010;
int _l[N * 4], _r[N * 4];
int _zero[N * 4], _one[N * 4];
int _lazy[N * 4];
int n, q;
char s[N], f[N];
int l[N], r[N];
bool flag;

void pushup(int pos)
{
    _zero[pos] = _zero[lson] + _zero[rson];
    _one[pos] = _one[lson] + _one[rson];
}

void pushdown(int pos)
{
    if (_lazy[pos] == -1)   return;
    else
    {
        if (_lazy[pos] == 1)
        {
            _one[lson] = _r[lson] - _l[lson] + 1;
            _zero[lson] = 0;
            _one[rson] = _r[rson] - _l[rson] + 1;
            _zero[rson] = 0;
        }
        else if (_lazy[pos] == 0)
        {
            _zero[lson] = _r[lson] - _l[lson] + 1;
            _one[lson] = 0;
            _zero[rson] = _r[rson] - _l[rson] + 1;
            _one[rson] = 0;
        }
        _lazy[lson] = _lazy[rson] = _lazy[pos];
        _lazy[pos] = -1;
    }
}

void build(int pos, int l, int r)
{
    _l[pos] = l, _r[pos] = r;
    _lazy[pos] = -1;
    if (l == r)
    {
        if (f[l] == '0')
        {
            _zero[pos] = 1;
            _one[pos] = 0;
        }
        else
        {
            _zero[pos] = 0;
            _one[pos] = 1;
        }
        return;
    }
    else
    {
        int mid = l + r >> 1;
        build(lson, l, mid);
        build(rson, mid + 1, r);
        pushup(pos);
    }
}


void modify(int pos, int l, int r, int x, int y, int tar)
{
    // all included
    if (x <= l && y >= r)
    {
        _lazy[pos] = tar;
        if (tar == 0)
            _zero[pos] = _r[pos] - _l[pos] + 1, _one[pos] = 0;
        else
            _one[pos] = _r[pos] - _l[pos] + 1, _zero[pos] = 0;
        return;
    }
    else
    {
        int mid = l + r >> 1;
        pushdown(pos);

        if (x <= mid)
        {
            modify(lson, l, mid, x, y, tar);
        }

        if (y >= mid + 1)
        {
            modify(rson, mid + 1, r, x, y, tar);
        }

        pushup(pos);
    }
}

void query(int pos, int l, int r, int x, int y, int &num0, int &num1)
{
    // all included
    if (x <= l && y >= r)
    {
        num0 = _zero[pos];
        num1 = _one[pos];
        return;
    }
    else
    {
        int mid = l + r >> 1;
        pushdown(pos);
        int tmp1, tmp2;
        num0 = num1 = 0;
        if (x <= mid)
        {
            query(lson, l, mid, x, y, tmp1, tmp2);
            num0 = tmp1, num1 = tmp2;
        }

        if (y >= mid + 1)
        {
            query(rson, mid + 1, r, x, y, tmp1, tmp2);
            num0 += tmp1, num1 += tmp2;
        }
        return;
    }
}



int main()
{
    int t; cin >> t;
    while (t -- )
    {
        // input
        scanf("%d%d", &n, &q);
        scanf("%s%s", s + 1, f + 1);
        for (int i = 1; i <= q; i ++ )
        {
           scanf("%d%d", &l[i], &r[i]);
        }

        // initial
        flag = true;
        build(1, 1, n);

        // opt
        for (int i = q; flag && i >= 1; i -- )
        {
            static int num0, num1;
            query(1, 1, n, l[i], r[i], num0, num1);
            if (num0 == num1)
            {
                flag = false;
                break;
            }
            else if (num0 > num1)   // 变成 0
            {
                modify(1, 1, n, l[i], r[i], 0);
                /// printf("(%d, %d)->%d\n", l[i], r[i], 0);
            }
            else    // 变成 1
            {
                modify(1, 1, n, l[i], r[i], 1);
                /// printf("(%d, %d)->%d\n", l[i], r[i], 1);
            }
        }
         // Check
        if (!flag)
        {
            // printf("因为之前的不行\n");
            puts("NO");
        }
        else
        {
            for (int i = 1; i <= n; i ++ )
            {
                static int num0, num1;
                query(1, 1, n, i, i, num0, num1);
                if (num0 == 1 && s[i] == '0')
                {
                    continue;
                }
                else if (num1 == 1 && s[i] == '1')
                {
                    continue;
                }
                else
                {
                    flag = false;
                    break;
                }
            }
            if (flag)
                puts("YES");
            else
                puts("NO");
        }

    }
    return 0;
}
/*
1
5 2
00000
00111
1 5
1 3
*/

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---