cf818A
1 #include <bits/stdc++.h>
2 #define ll __int64
3 using namespace std;
4
5 ll n, k;
6
7 int main() {
8 cin >> n >> k;
9 ll a = n / 2 / (k+1);
10 ll b = k * a;
11 printf("%I64d %I64d %I64d\n", a, b, n-a-b);
12 return 0;
13 }
818B，没有指示的数字随便赋值即可。
1 #include <bits/stdc++.h>
2 #define ll __int64
3 using namespace std;
4
5 int n, m;
6 int l[105];
7 int a[105];
8 bool mark[105];
9 vector<int> v;
10
11 int main() {
12 cin >> n >> m;
13 for (int i = 0; i < m; i++) {
14 cin >> l[i];
15 int k = (l[i] - l[i-1] + n) % n;
16 if (!k) k = n;
17 if (!a[l[i-1]]) {
18 a[l[i-1]] = k;
19 } else if (a[l[i-1]] != k) {
20 puts("-1");
21 return 0;
22 }
23 }
24
25 for (int i = 1; i <= n; i++) {
26 if (a[i]) {
27 if (!mark[a[i]])
28 mark[a[i]] = true;
29 else {
30 puts("-1");
31 return 0;
32 }
33 } else {
34 v.push_back(i);
35 }
36 }
37 for (int i = 1; i <= n; i++) {
38 if (!mark[i]) {
39 a[v.back()] = i;
40 v.pop_back();
41 }
42 }
43
44 for_each(a+1, a+1+n, [](int x) {cout << x << " ";});
45 return 0;
46 }
818C，sweep line，坐标点为下标差分计数。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxcor = 1e5 + 5;
5 int d, n, m;
6 int cntl, cntr, cntt, cntb;
7 struct Point {
8 int x1, y1, x2, y2;
9 }sofa[maxcor];
10 int cnt_left[maxcor], cnt_right[maxcor], cnt_top[maxcor], cnt_bottom[maxcor];
11
12 inline bool l(int i) {
13 int k = cnt_left[max(sofa[i].x1, sofa[i].x2) - 1];
14 if (sofa[i].x1 != sofa[i].x2) k--;
15 return k == cntl;
16 }
17
18 inline bool r(int i) {
19 int k = cnt_right[min(sofa[i].x1, sofa[i].x2) + 1];
20 if (sofa[i].x1 != sofa[i].x2) k--;
21 return k == cntr;
22 }
23
24 inline bool t(int i) {
25 int k = cnt_top[max(sofa[i].y1, sofa[i].y2) - 1];
26 if (sofa[i].y1 != sofa[i].y2) k--;
27 return k == cntt;
28 }
29
30 inline bool b(int i) {
31 int k = cnt_bottom[min(sofa[i].y1, sofa[i].y2) + 1];
32 if (sofa[i].y1 != sofa[i].y2) k--;
33 return k == cntb;
34 }
35
36 int main() {
37 cin >> d >> n >> m;
38 for (int i = 1; i <= d; i++) {
39 int x1, y1, x2, y2;
40 cin >> x1 >> y1 >> x2 >> y2;
41 sofa[i] = {x1, y1, x2, y2};
42
43 cnt_left[min(x1, x2)]++;
44 cnt_right[max(x1, x2)]++;
45 cnt_top[min(y1, y2)]++;
46 cnt_bottom[max(y1, y2)]++;
47 }
48 cin >> cntl >> cntr >> cntt >> cntb;
49
50 for (int i = 1; i <= 1e5; i++) {
51 cnt_left[i] += cnt_left[i-1];
52 cnt_top[i] += cnt_top[i-1];
53 }
54 for (int i = 1e5; i; i--) {
55 cnt_right[i] += cnt_right[i+1];
56 cnt_bottom[i] += cnt_bottom[i+1];
57 }
58
59 for (int i = 1; i <= d; i++) {
60 if (l(i) && r(i) && t(i) && b(i)) {
61 cout << i << endl;
62 return 0;
63 }
64 }
65 puts("-1");
66 return 0;
67 }
818D，gameover之时将此数打标记，否则继续计数。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxcolor = 1e6 + 5;
5 bool mark[maxcolor];
6 int n, A, cntA;
7 int cnt[maxcolor];
8
9 int main() {
10 cin >> n >> A;
11 mark[A] = true;
12 for (int i = 1; i <= n; i++) {
13 int a;
14 scanf("%d", &a);
15 if (a == A) cntA++;
16 else {
17 if (mark[a])
18 continue;
19 else if (cnt[a] < cntA)
20 mark[a] = true;
21 else cnt[a]++;
22 }
23 }
24 for (int i = 1; i <= 1e6; i++)
25 if (!mark[i] && cnt[i] >= cntA) {
26 cout << i << endl;
27 return 0;
28 }
29 puts("-1");
30 return 0;
31 }
BZOJ1011，鬼题勿做。思路根本就不是在求正解，以及迷之SPJ。
1 #include <cstdio>
2 #define db double
3
4 const int maxn = 1e5 + 5;
5 int n;
6 db A;
7 db m[maxn], sum[maxn];
8
9 int main() {
10 scanf("%d%lf", &n, &A);
11 for (int i = 1; i <= n; i++) {
12 scanf("%lf", &m[i]);
13 sum[i] = sum[i-1] + m[i];
14 }
15 for (int i = 1; i <= n; i++) {
16 int k = int(A * db(i) + 1e-8);
17 db ans = 0.0;
18 if (k <= 100) {
19 for (int j = 1; j <= k; j++)
20 ans += m[i] * m[j] / (i-j);
21 } else {
22 ans = sum[k] * m[i] / (i-k/2);
23 }
24 printf("%.6lf\n", ans);
25 }
26 return 0;
27 }
POJ3468，分块。
1 #include <cstdio>
2 #include <cmath>
3 #define ll long long
4 #define maxn 100005
5 #define rep(i, x, y) for (int i = x; i <= y; i++)
6
7 int n, m, t;
8 int L[maxn], R[maxn], pos[maxn];
9 ll a[maxn];
10 ll sum[maxn], add[maxn];
11
12 void Add(int l, int r, int k) {
13 int p = pos[l], q = pos[r];
14 if (p == q) {
15 rep(i, l, r) a[i] += k;
16 sum[p] += k * (r - l + 1);
17 } else {
18 rep(i, p+1, q-1) add[i] += k;
19 rep(i, l, R[p]) a[i] += k;
20 sum[p] += k * (R[p] - l + 1);
21 rep(i, L[q], r) a[i] += k;
22 sum[q] += k * (r - L[q] + 1);
23 }
24 }
25
26 ll query(int l, int r) {
27 int p = pos[l], q = pos[r];
28 ll ans = 0ll;
29 if (p == q) {
30 rep(i, l, r) ans += a[i];
31 ans += add[p] * (r - l + 1);
32 } else {
33 rep(i, p+1, q-1) ans += sum[i] + add[i] * (R[i] - L[i] + 1);
34 rep(i, l, R[p]) ans += a[i];
35 ans += add[p] * (R[p] - l + 1);
36 rep(i, L[q], r) ans += a[i];
37 ans += add[q] * (r - L[q] + 1);
38 }
39 return ans;
40 }
41
42 int main() {
43 scanf("%d%d", &n, &m);
44 rep(i, 1, n) scanf("%lld", &a[i]);
45
46 t = sqrt(n);
47 rep(i, 1, t) {
48 L[i] = (i-1) * t + 1;
49 R[i] = i * t;
50 }
51 if (R[t] < n) {
52 t++;
53 L[t] = R[t-1] + 1;
54 R[t] = n;
55 }
56 rep(i, 1, t) {
57 rep(j, L[i], R[i]) {
58 pos[j] = i;
59 sum[i] += a[j];
60 }
61 }
62
63 rep(i, 1, m) {
64 char ch[3];
65 int a, b;
66 scanf("%s%d%d", ch, &a, &b);
67 if (ch[0] == ‘C‘) {
68 int c;
69 scanf("%d", &c);
70 Add(a, b, c);
71 } else {
72 printf("%lld\n", query(a, b));
73 }
74 }
75
76 return 0;
77 }
原文地址：https://www.cnblogs.com/AlphaWA/p/10363662.html
时间： 2024-10-30 03:45:23