Codeforces Round #519
題目連結:傳送門
A. Elections (思維+暴力)
思路:
從最小的k開始列舉就好了- -。
#include <bits/stdc++.h> using namespace std; const int MAX_N = 100 + 5; int a[MAX_N]; int main() { int N; cin >> N; int m = -1, sum = 0; for (int i = 1; i <= N; i++) { scanf("%dView Code", a+i); m = max(m, a[i]); sum += a[i]; } int ans = m; int Awruk = N*m-sum; while (Awruk <= sum) { Awruk += N; ans++; } cout << ans << endl; return 0; }
codeforces1043B. Lost Array (暴力列舉)
思路:
把差分預處理出來,列舉長度,可以跑到最後的就加入答案。
#include <bits/stdc++.h> using namespace std; const int MAX_N = 1e3 + 5; int a[MAX_N]; int b[MAX_N]; int ans[MAX_N]; int main() { int N; cin >> N; a[0] = 0; for (int i = 1; i <= N; i++) { scanf("%d", a+i); b[i]View Code= a[i] - a[i-1]; } int cnt = 0; for (int i = 1; i <= N; i++) { int j; for (j = 1; j <= N; j++) { if (b[j] != b[(j-1)%i+1]) break; } if (j == N+1) ans[cnt++] = i; } cout << cnt << endl << ans[0]; for (int i = 1; i < cnt; i++) cout << ' ' << ans[i]; cout << endl; return 0; }
codeforces1043C. Smallest Word (思維)
思路:
遇到連續的a就找到最後一個,翻轉。
遇到連續的b也找到最後一個,如果下一個是a,翻轉。
#include <bits/stdc++.h> using namespace std; const int MAX_N = 1e3 + 5; bool ans[MAX_N]; int main() { string s; memset(ans, false, sizeof ans); cin >> s; int len = s.size(); int i = 0; while (i < len) { while (i < len && s[i] == 'a') i++; ans[i-1] = true; while (i < len && s[i] == 'b') i++; if (i < len && s[i] == 'a') ans[i-1] = true; } cout << ans[0]; for (int i = 1; i < len; i++) cout << ' ' << ans[i]; cout << endl; return 0; }View Code
codeforces1043D. Mysterious Crime (尺取+暴力)
思路:
從第一行開始,往下面幾行找最長的匹配。每行從上一行的最左端的數開始往右匹配,匹配到不同的就break,進入下一行。
預處理一個pos[i][j],表示第i行的值為j的數在哪個位置。
找完最後一行後得到最長的匹配的長度為len,計算len對答案產生的貢獻。
注意M為1的時候不能匹配要特判一下,答案就是N*(N+1)/2。
#include <bits/stdc++.h> using namespace std; const int MAX_N = 1e5 + 5; const int MAX_M = 10 + 3; int N, M; int a[MAX_M][MAX_N]; int pos[MAX_M][MAX_N]; bool vis[MAX_N]; int main() { cin >> N >> M; memset(vis, false, sizeof vis); for (int i = 1; i <= M; i++) for (int j = 1; j <= N; j++) { scanf("%d", &a[i][j]); pos[i][a[i][j]] = j; } long long ans = 0; for (int i = 1; i <= N; i++) if (!vis[a[1][i]]){ int l = i, r = N; long long len = 1; for (int j = 2; j <= M; j++) { int nxtl, nxtr; nxtl = nxtr = pos[j][a[j-1][l]]; while (l <= r && nxtr <= N && a[j][nxtr] == a[j-1][l]) { l++; nxtr++; } l = nxtl, r = nxtr-1; if (j == M) { len = r-l+1; for (int k = l; k <= r; k++) vis[a[M][k]] = true; } } ans += (len+1)*len/2; } if (M == 1) ans = (long long)(N+1)*N/2; cout << ans << endl; return 0; }View Code
codeforces1043E. Train Hard, Win Easy(思維+排序)
思路:
兩個人(u,v)組隊時他們的得分是min(xu+yv,xv+yu),考慮到:
xu+yv < xv+yu
=>yv-xv < yu-xu
令di = yi-xi,則組隊時d小的一方出y,d大的一方出x。
根據d排序後,對於(1 ≤ i < j ≤ N)有di < dj,所以第i個人和第j個人組隊的得分為yi+xj。
預處理x和y的字首和,對於第i個人,他與在他之前的人組隊的得分為:
$\sum_{j=1}^{i-1}y_{j}$ + xi * (i-1);
與在他之後的人組隊的得分為:
$\sum_{j=i+1}^{N}x_{j}$ + yi * (N-i);
這樣可以O(1)算出每個人的得分,然後再減去不能和自己比賽的人的貢獻就好了。
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int MAX_N = 3e5 + 5; struct Node{ int ind; ll x, y; ll d; Node(int _i = 0, ll _x = 0, ll _y = 0) : ind(_i), x(_x), y(_y) {d = y-x;} bool operator < (const Node& a) const { return d < a.d; } }nodes[MAX_N]; ll sumx[MAX_N], sumy[MAX_N]; ll ans[MAX_N]; int main() { int N, M; cin >> N >> M; memset(ans, 0, sizeof ans); for (int i = 1; i <= N; i++) { ll x, y; scanf("%I64d%I64d", &x, &y); nodes[i] = Node(i, x, y); } for (int i = 1; i <= M; i++) { int u, v; scanf("%d%d", &u, &v); ll tmp = 0; if (nodes[u] < nodes[v]) tmp = nodes[u].y + nodes[v].x; else tmp = nodes[v].y + nodes[u].x; ans[u] -= tmp; ans[v] -= tmp; } sort(nodes+1, nodes+1+N); sumx[0] = sumy[0] = 0; for (int i = 1; i <= N; i++) { sumx[i] = sumx[i-1] + nodes[i].x; sumy[i] = sumy[i-1] + nodes[i].y; } for (int i = 1; i <= N; i++) { Node cur = nodes[i]; ans[cur.ind] += sumy[i-1] + cur.x*(i-1); ans[cur.ind] += sumx[N] - sumx[i] + cur.y*(N-i); } bool firstprint = true; for (int i = 1; i <= N; i++) { if (firstprint) firstprint = false; else printf(" "); printf("%I64d", ans[i]); } puts(""); return 0; }View Code
codeforces1043F. Make It One(組合數學+數論+dp)
思路:
說實話這題比較玄學。
2 * 3 * 5 * 7 * 11 * 13 * 17 ≈ 5e5,所以3e5範圍內一個數最多有6個質因數。所以答案是不會超過6的,考慮從1開始列舉答案。
#include <bits/stdc++.h> using namespace std; const int MAX_N = 3e5 + 5; const int MOD = 1e9 + 7; int fpow(int a, int p) { int ans = 1; while (p) { if (p&1) ans = (1LL * ans * a) % MOD; p >>= 1; a = (1LL * a * a) % MOD; } return ans; } int N; int a[MAX_N]; int f[20][MAX_N], cnt[MAX_N]; int fac[MAX_N], inv[MAX_N]; int newton(int m, int n) { if (n < 0 || m < n) return 0; return ((1LL * fac[m] * inv[n])%MOD * inv[m-n]) % MOD; } void init() { fac[0] = 1; for (int i = 1; i <= MAX_N; i++) fac[i] = (1LL * fac[i-1] * i) % MOD; inv[MAX_N-1] = fpow(fac[MAX_N-1], MOD-2); for (int i = MAX_N-1; i >= 1; i--) inv[i-1] = (1LL * inv[i] * i) % MOD; } void sub(int& a, int b) { a -= b; if (a < 0) a += MOD; } int main() { init(); scanf("%d", &N); memset(cnt, 0, sizeof cnt); for (int i = 1; i <= N; i++) { scanf("%d", a+i); cnt[a[i]]++; } for (int i = 1; i < MAX_N; i++) { for (int j = 2*i; j < MAX_N; j += i) { cnt[i] += cnt[j]; } } for (int i = 1; i < 20; i++) { for (int j = MAX_N-1; j >= 1; j--) { f[i][j] = newton(cnt[j], i); for (int k = j+j; k < MAX_N; k += j) sub(f[i][j], f[i][k]); } if (f[i][1] > 0) { printf("%d\n", i); return 0; } } puts("-1"); return 0; }View Code