A：后缀维护si*pi的最小值，查询的时候二分，判断后缀和当前两个部分就行。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 typedef long long LL;
5 const int maxn = 100100;
6 int n, m;
7 LL s[maxn], p[maxn];
8 LL suf[maxn];
9
10 int main() {
11 // freopen("in", "r", stdin);
12 int T;
13 LL q;
14 scanf("%d", &T);
15 while(T--) {
16 scanf("%d%d",&n,&m);
17 memset(suf, 0, sizeof(suf));
18 for(int i = 0; i < n; i++) scanf("%lld%lld",&s[i],&p[i]);
19 suf[n-1] = s[n-1] * p[n-1];
20 for(int i = n - 2; i >= 0; i--) suf[i] = min(suf[i+1], s[i]*p[i]);
21 s[n] = 100000000000000LL; suf[n] = 100000000000000LL;
22 while(m--) {
23 scanf("%lld", &q);
24 int pos = lower_bound(s, s+n+1, q) - s;
25 // if(s[pos] == q) printf("%lld\n", q*p[pos]);
26 if(pos == n) printf("%lld\n", q*p[n-1]);
27 else printf("%lld\n", min(suf[pos], q*p[pos-1]));
28 }
29 }
30 return 0;
31 }
B：先离散化找出所有内容的种类，之后尺取卡出最小页数。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxn = 1001000;
5 int n, m, rl, rr;
6 int h[maxn], p[maxn];
7 int vis[maxn];
8 int cnt, ret;
9
10 int id(int x) {
11 return lower_bound(h, h+m, x) - h;
12 }
13
14 int main() {
15 // freopen("in", "r", stdin);
16 while(~scanf("%d", &n)) {
17 for(int i = 0; i < n; i++) {
18 scanf("%d", &p[i]);
19 h[i] = p[i];
20 }
21 sort(h, h+n); m = unique(h, h+n) - h;
22 memset(vis, 0, sizeof(vis));
23 cnt = 0; ret = n; rl = 0, rr = n - 1;
24 int lo = 0, hi = 0;
25 while(lo < n) {
26 while(hi < n && cnt < m) {
27 int idx = id(p[hi++]);
28 if(!vis[idx]) cnt++;
29 vis[idx]++;
30 }
31 if(cnt < m) break;
32 if(ret > hi - lo && cnt == m) {
33 ret = hi - lo;
34 rl = lo, rr = hi - 1;
35 }
36 // printf("%d %d\n", lo, hi);
37 int idx = id(p[lo++]);
38 vis[idx]--;
39 if(vis[idx] == 0) cnt--;
40 if(lo >= n) break;
41 }
42 printf("%d\n", ret);
43 printf("%d %d\n", rl, rr);
44 }
45 return 0;
46 }
C：水题…
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxn = 100100;
5 int n;
6 int a[maxn], b[maxn];
7
8 int main() {
9 // freopen("in", "r", stdin);
10 int T;
11 scanf("%d", &T);
12 while(T--) {
13 scanf("%d", &n);
14 int x = 0, y = 0;
15 for(int i = 1; i <= n; i++) {
16 scanf("%d%d",&a[i],&b[i]);
17 }
18 sort(a+1, a+n+1);
19 sort(b+1, b+n+1);
20 for(int i = 1; i <= n; i++) {
21 if(a[i] > b[i]) x += 2;
22 else if(a[i] == b[i]) x++, y++;
23 else y += 2;
24 }
25 if(x > y) puts("Winjourn");
26 else if(x == y) puts("Tie");
27 else puts("Never_Sorry");
28 }
29 return 0;
30 }
D：数位DP，dp(l,cnt)统计长度为l到时候1的个数为cnt的整数。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 typedef long long LL;
5 const int maxn = 22;
6 int digit[maxn];
7 LL dp[maxn][maxn];
8 LL n;
9
10 LL dfs(int l, int cnt, bool flag) {
11 if(l == 0) return cnt;
12 if(!flag && ~dp[l][cnt]) return dp[l][cnt];
13 int pos = flag ? digit[l] : 9;
14 LL ret = 0;
15 for(int i = 0; i <= pos; i++) {
16 ret += dfs(l-1, cnt+(i==1), flag&&(pos==i));
17 }
18 if(!flag) dp[l][cnt] = ret;
19 return ret;
20 }
21
22 LL f(LL x) {
23 int pos = 0;
24 while(x) {
25 digit[++pos] = x % 10;
26 x /= 10;
27 }
28 return dfs(pos, 0, true);
29 }
30
31 int main() {
32 // freopen("in", "r", stdin);
33 int _ = 1;
34 memset(dp, -1, sizeof(dp));
35 while(~scanf("%lld", &n)) {
36 printf("Case %d:%lld\n", _++, f(n));
37 }
38 return 0;
39 }
E：模拟，如果循环节正好长度为50的话，就不要打省略号了。。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 vector<int> t1, t2;
5 map<int, int> id;
6 int a, b;
7
8 void f(int a, int b) {
9 id.clear();
10 t1.clear(); t2.clear();
11 int p = -1;
12 while(1) {
13 t1.push_back(a/b);
14 int x = a % b;
15 if(x == 0) break;
16 if(id.find(x) == id.end()) {
17 id[x] = t2.size();
18 t2.push_back(x);
19 }
20 else {
21 p = id[x];
22 break;
23 }
24 a = x * 10;
25 }
26 if(t1.size() == 0) while(1);
27 printf("%d.", t1[0]);
28 bool flag = 0;
29 bool plusflag = 0;
30 int cnt = 0;
31 for(int i = 1; i < t1.size(); i++) {
32 if(cnt >= 50) {
33 plusflag = 1;
34 break;
35 }
36 if(i == p + 1) {
37 printf("(");
38 flag = 1;
39 }
40 printf("%d", t1[i]);
41 if(flag) cnt++;
42 }
43 if(plusflag) printf("...");
44 if(p >= 0) {
45 if(flag) printf(")\n");
46 printf("%d\n", t1.size()-p-1);
47 }
48 else {
49 printf("(0)\n");
50 printf("1\n");
51 }
52 }
53
54 int main() {
55 // freopen("in", "r", stdin);
56 // freopen("out", "w", stdout);
57 int _ = 1;
58 while(~scanf("%d%d",&a,&b)) {
59 printf("Case %d:\n%d/%d=", _++, a, b);
60 if(a % b == 0) printf("%d.(0)\n1\n", a / b);
61 else f(a, b);
62 }
63 return 0;
64 }
F：好麻烦的计算几何，感觉要先卡出封闭图形，之后找到所有的矩形。然后看看能放的最大圆。
G：对整个数组排序后二分答案，贪心地从左到右选取数字，保证相邻两个数的差不小于二分出来的值，判断是否满足大于等于k个。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxn = 100100;
5 int n, k;
6 int a[maxn], s[maxn];
7 int ret;
8
9 bool ok(int dis) {
10 int i = 2, p = 1;
11 int cnt = 0;
12 while(i <= n) {
13 if(a[i] - a[p] >= dis) {
14 p = i; i++;
15 cnt++;
16 }
17 else i++;
18 }
19 return cnt + 1 >= k;
20 }
21
22 int main() {
23 // freopen("in", "r", stdin);
24 while(~scanf("%d%d",&n,&k)) {
25 memset(s, 0, sizeof(s));
26 for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
27 sort(a+1, a+n+1);
28 ret = a[n] - a[1];
29 int lo = 0, hi = a[n];
30 while(lo <= hi) {
31 int mid = (lo + hi) >> 1;
32 if(ok(mid)) {
33 ret = mid;
34 lo = mid + 1;
35 }
36 else hi = mid - 1;
37 }
38 printf("%d\n", ret);
39 }
40 return 0;
41 }
H：像是需要用启发式搜索将整座附近的点处理出来一个优先值，顺着第一条路走，最后回来。（后来有巨巨说是BFS）
I：枚举a,b字符串子串的起始，扫描对应位，如果不相同则替换至相同，最多k次操作。贪心地将替换操作放到前面，如果不能替换则break更新ret。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <string>
6 #include <map>
7 using namespace std;
8
9 const int maxn = 550;
10 int n, m, k;
11 char a[maxn], b[maxn];
12
13 int main() {
14 // freopen("in" ,"r", stdin);
15 while(~scanf("%d", &k)) {
16 scanf("%s%s",a,b);
17 n = strlen(a); m = strlen(b);
18 int ret = 0;
19 for(int i = 0; i < n; i++) {
20 for(int j = 0; j < m; j++) {
21 int cnt = 0, tmp = 0;
22 for(int aa = i, bb = j; aa < n && bb < m; aa++, bb++) {
23 if(a[aa] == b[bb]) tmp++;
24 else {
25 if(cnt < k) tmp++, cnt++;
26 else break;
27 }
28 }
29 ret = max(ret, tmp);
30 }
31 }
32 printf("%d\n", ret);
33 }
34 return 0;
35 }
J：f(i,j)表示和为i，用j个数字组成的方案数，转移方程f(i,j)=f(i-p,j-1)，p取1~4，会爆LL所以用double存了f。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 typedef long long LL;
5 const int maxn = 111;
6 const int maxm = 666;
7 int n, m;
8 double f[maxn][maxm];
9
10 int main() {
11 // freopen("in", "r", stdin);
12 memset(f, 0, sizeof(f));
13 f[1][1] = 1.0; f[1][2] = 1.0; f[1][3] = 1.0; f[1][4] = 1.0;
14 for(int i = 2; i < maxn; i++) {
15 for(int j = 1; j < maxm; j++) {
16 for(int k = 1; k <= 4; k++) {
17 if(k >= j) continue;
18 f[i][j] += f[i-1][j-k];
19 }
20 }
21 }
22 while(~scanf("%d%d",&n,&m)) {
23 printf("%.4lf\n", f[n][m]/pow(4,n));
24 }
25 return 0;
26 }
K：最后的情形无非是两种，一种是wywy..，另一种是ywyw..。整个串长度如果是奇数那么是无法围成要求的项链的；之后判断当前情况和上述哪种情况最近，然后分别找w和y错位的数量，先交换，再修改，min(w,y)+(max(w,y)-min(w,y))=max(w,y)即为答案。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxn = 100100;
5 char s[maxn], t[2][maxn];
6 int n;
7
8 int main() {
9 // freopen("in", "r", stdin);
10 memset(t, 0, sizeof(t));
11 for(int i = 0; i < maxn; i++) {
12 if(i % 2 == 0) {
13 t[0][i] = ‘w‘, t[1][i] = ‘y‘;
14 }
15 else {
16 t[0][i] = ‘y‘, t[1][i] = ‘w‘;
17 }
18 }
19 while(~scanf("%d", &n)) {
20 scanf("%s", s);
21 if(n & 1) {
22 puts("What a pity!");
23 continue;
24 }
25 int a = 0, b = 0;
26 for(int i = 0; i < n; i++) {
27 if(s[i] != t[0][i]) a++;
28 if(s[i] != t[1][i]) b++;
29 }
30 if(a < b) {
31 int w = 0, y = 0;
32 for(int i = 0; i < n; i++) {
33 if(s[i] != t[0][i]) {
34 if(s[i] == ‘w‘) w++;
35 else y++;
36 }
37 }
38 printf("%d\n", max(w, y));
39 }
40 else {
41 int w = 0, y = 0;
42 for(int i = 0; i < n; i++) {
43 if(s[i] != t[1][i]) {
44 if(s[i] == ‘w‘) w++;
45 else y++;
46 }
47 }
48 printf("%d\n", max(w, y));
49 }
50 }
51 return 0;
52 }
L：不会…
时间： 2024-12-25 09:54:44