Codeforces Round #697 (Div.3) A~G解题报告与解法证明

题目大体概括

A

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;typedef long long LL;
const int N = 500;
LL a[N];
int sz;
bool Check(LL n)
{for (int i = 0; i <= sz; i ++ ){if (n == a[i]) return false;}return true;
}int main()
{int T;  cin >> T;LL n;a[0] = 1;sz = 60;for (int i = 1; i <= sz; i ++ )a[i] = a[i - 1] * 2;while (T -- ){scanf("%lld", &n);if (Check(n)){cout << "YES" << endl;}else{cout << "NO" << endl;}}return 0;
}

B

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;typedef long long LL;
const int N = 1000010;
bool st[N + 10];int main()
{int T;  cin >> T;memset(st, false, sizeof st);st[2020] = st[2021] = true;for (int i = 2022; i < N; i ++ ){st[i] = st[i - 2020] | st[i - 2021];}while (T -- ){static int x;   scanf("%d", &x);if (st[x])  cout << "YES\n";else    cout << "NO\n";}return 0;
}

C

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;typedef long long LL;
const int N = 200010;
int cnt1[N], cnt2[N];
int n, m, k;int main()
{int T;  cin >> T;while (T -- ){scanf("%d%d%d", &n, &m, &k);static int x;   static LL res;res = 0LL;memset(cnt1, 0, sizeof cnt1);memset(cnt2, 0, sizeof cnt2);for (int i = 1; i <= k; i ++ ){scanf("%d", &x);cnt1[x] ++;}for (int i = 1; i <= k; i ++ ){scanf("%d", &x);cnt2[x] ++;}res = LL(k) * (k - 1);for (int i = 1; i <= n; i ++ ){res -= LL(cnt1[i]) * (cnt1[i] - 1);}for (int i = 1; i <= m; i ++ ){res -= LL(cnt2[i]) * (cnt2[i] - 1);}cout << res / 2 << endl;}return 0;
}

D

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;typedef long long LL;
const int N = 200010, INF = 0x3f3f3f3f;
int n, m;
int a[N];
LL c[N], d[N];
int szc, szd;bool cmp(LL a, LL b)
{return a > b;
}
int main()
{int T;  cin >> T;while (T -- ){static int b;cin >> n >> m;for (int i = 1; i <= n; i ++ )  scanf("%d", &a[i]);c[0] = d[0] = szc = szd = 0;for (int i = 1; i <= n; i ++ ){scanf("%d", &b);if (b & 1)c[++ szc] = a[i];elsed[++ szd] = a[i];}sort(c + 1, c + szc + 1, cmp);sort(d + 1, d + szd + 1, cmp);for (int i = 1; i <= szc; i ++ )    c[i] += c[i - 1];for (int i = 1; i <= szd; i ++ )    d[i] += d[i - 1];if (c[szc] + d[szd] < m){puts("-1");}else{int res = INF;for (int i = 0; i <= szc; i ++ ){if (c[i] + d[szd] < m)continue;else if (c[i] >= m){res = min(res, i);break;}else{static int tmp;tmp = lower_bound(d + 1, d + 1 + szd, m - c[i]) - d;res = min(res, tmp + tmp + i);}}cout << res << endl;}}return 0;
}

E

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;typedef long long LL;
const int N = 1010, INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n, k;
int a[N];
int f[N][N];bool cmp(LL a, LL b)
{return a > b;
}int main()
{memset(f, 0, sizeof f);for (int i = 0; i < N; i ++ )f[i][i] = f[i][0] = 1;for (int i = 1; i < N; i ++ )for (int j = 1; j <= i; j ++ )f[i][j] = (f[i - 1][j] + f[i - 1][j - 1]) % MOD;int T;  cin >> T;while (T -- ){cin >> n >> k;for (int i = 1; i <= n; i ++ )scanf("%d", &a[i]);sort(a + 1, a + n + 1, cmp);static int x, sidx, eidx;x = a[k], sidx = -1, eidx = -1;for (int i = 1; i <= n; i ++ ){if (a[i] == x){if (sidx == -1) sidx = i;eidx = i;}}cout << f[eidx - sidx + 1][k - sidx + 1] << endl;}return 0;
}

F

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;typedef long long LL;
const int N = 1010, INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n;
char a[N][N], b[N][N];
char c[N][N];inline void Change(char c[][N], int x, bool row);
bool Same(char c[][N], char d[][N], int x)
{for (int i = 1; i <= n; i ++ )if (c[x][i] != d[x][i])return false;return true;
}bool Check(char c[][N], char d[][N])
{// 第一行是不需要动的，我们看看 列 的影响for (int i = 1; i <= n; i ++ )if (c[1][i] != d[1][i])Change(c, i, false);for (int i = 2; i <= n; i ++ ){if (c[i][1] != d[i][1]) // 修改 行Change(c, i, true);if (!Same(c, d, i))return false;}return true;
}
inline void Change(char c[][N], int x, bool row)
{if (row)    // rowfor (int i = 1; i <= n; i ++ ){c[x][i] = 97 - c[x][i];   // 48 + 49 - c[i]}else        // colfor (int i = 1; i <= n; i ++ ){c[i][x] = 97 - c[i][x];   // 48 + 49 - c[i]}
}int main()
{int T;  cin >> T;while (T -- ){cin >> n;for (int i = 1; i <= n; i ++ )scanf("%s", a[i] + 1);for (int i = 1; i <= n; i ++ )scanf("%s", b[i] + 1);memcpy(c, a, sizeof c);if (Check(c, b) || (memcpy(c, a, sizeof c), Change(c, 1, true), Check(c, b)))puts("YES");elseputs("NO");}return 0;
}

G

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;typedef long long LL;
const int N = 200010, INF = 0x3f3f3f3f;int cnt[N];
int a[N];
bool st[N];
int f[N];
int n;int main()
{int T;  cin >> T;while (T -- ){static int res;cin >> n;for (int i = 1; i <= n; i ++ )scanf("%d", &a[i]);res = INF;sort(a + 1, a + 1 + n);memset(cnt, 0, sizeof cnt);memset(st, false, sizeof st);memset(f, 0, sizeof f);for (int i = 1; i <= n; i ++ )  cnt[a[i]] ++;for (int i = 1, val; i <= n; i ++ ){val = a[i];if (st[val])    continue;st[val] = true;// cnt[val] = max(cnt[val], 1);f[val] = f[val] + cnt[val];    // 给自己加的for (int j = val + val; j < N; j += val){f[j] = max(f[j], f[val]); // 给别的数字加的}res = min(res, n - f[val]);}cout << res << endl;}return 0;
}

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Codeforces Round #697 (Div.3) A~G解题报告与解法证明

题目大体概括

A

B

C

D

E

F

G

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