以下题解包括：\(A \ \ \ B \ \ \ C \ \ \ D \ \ \ E \ \ \ J\)

另：\(H【Pair】数位dp \ 待补\) 毕竟我还不会

比赛地址： https://ac.nowcoder.com/acm/contest/887#question

【A】 String 最小表示法

给定一个字符串，要把它分成最少段的最小表示法串，输出分段之后的字符串。

从前往后依次枚举，不断缩短长度直到当前前缀为最小表示法即可，然后输出再从当前末尾开始判断。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;char s[205];
string temp;
int l;int get_min() {int i = 0, j = 1;int k = 0;while(i < l && j < l && k < l) {int t = temp[(i+k)%l] - temp[(j+k)%l];if(t == 0) {k ++;}else {if(t > 0) {i = i + k + 1;}else {j = j + k + 1;}k = 0;if(i == j) {j ++;}}}return i < j ? i : j;
}int main() {int t;scanf("%d", &t);while(t--) {scanf("%s", s);int n = strlen(s);int pos = 0;while(1) {temp = "";for(int i = pos; i < n; i++) {temp = temp + s[i];}l = temp.length();while(get_min() != 0) {l--;}pos = pos + l;for(int i = 0; i < l; i ++) {cout << temp[i];}if(pos == n) {printf("\n");break;}else {printf(" ");}}}return 0;
}

【B】 Irreducible Polynomial 数学

给定一个一元 \(n\) 次方程，问能不能把它分解成连乘的形式。

又是个定理题，次数大于等于 3 一定不行，等于 1 一定可以，等于二的时候需要判断 \(b^2 - 4ac < 0\) 是否成立，若成立，则可以。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;int a[30];int main() {0;
}int t;scanf("%d", &t);while(t--) {int n;scanf("%d", &n);for(int i = n; i >= 0; i--) {scanf("%d", &a[i]);}if(n >= 3) {printf("No\n");continue;}if(n == 1) {printf("Yes\n");continue;}if(a[1]*a[1]-4*a[0]*a[2] < 0) {printf("Yes\n");}else {printf("No\n");}}return 0;
}

【C】 Governing sand 贪心

我也不知道为什么比赛时候没看到 "c <= 200" 这个条件，然后队友敲了个权值线段树维护，虽然过了也是浪费了挺多时间的，还把学弟坑了...以下是赛后写的正解（大概）。

给定 \(n\) 种树，每种都有它的高度 \(h(\leq1e^9)\)，砍倒的费用 \(c(\leq 200)\)以及数量 \(p(\leq 1e^9)\)。现在需要让最高的树的数量占比超过 50%，问最小花费是多少。

其实就是各种排序贪心一下就好了。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;const int maxn = 1e5+5;int n;
ll sum[maxn];
ll cost[205];
ll numh[maxn];
vector<ll> v;
struct node {ll h, c, p;bool operator < (const node &q) const {return h < q.h;}
}a[maxn];int getid(ll x) {return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}int main() {while(~scanf("%d", &n)) {v.clear();for(int i = 1; i <= n; i++) {scanf("%lld%lld%lld", &a[i].h, &a[i].c, &a[i].p);v.push_back(a[i].h);}sort(a+1, a+1+n);sort(v.begin(), v.end());v.erase(unique(v.begin(), v.end()), v.end());memset(sum, 0, sizeof(sum));memset(numh, 0, sizeof(numh));for(int i = 1; i <= n; i++) {int id = getid(a[i].h);sum[id] += sum[id-1] + 1ll*a[i].p*a[i].c;numh[id] += numh[id-1] + 1ll*a[i].p;}ll ans = 1e18;memset(cost, 0, sizeof(cost));int new_n = v.size();for(int i = 1; i <= n; i++) {int id = getid(a[i].h);ll temp = sum[new_n] - sum[id]; // 删掉所有比它高的树的代价ll num = numh[id] - numh[id-1]; // 这种高度的树的数量ll num_small = numh[id-1];      // 比他矮的树的数量// cout << temp << "  " << num << "  " << num_small << endl;if(num > num_small) {           // 不用砍了ans = min(ans, temp);   cost[a[i].c] += 1ll*a[i].p;continue;}ll more = num_small - num + 1;  // 需要砍掉多少for(int j = 1; j <= 200; j++) {if(cost[j] == 0) {continue;}if(more - cost[j] >= 0) {more -= cost[j];temp = temp + cost[j]*j;}else {temp = temp + more*j;more = 0;}if(more == 0) {break;}}cost[a[i].c] += 1ll*a[i].p;ans = min(ans, temp);}printf("%lld\n", ans);}return 0;
}

【D】 Number 水题

水题不写了。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;int get(int x) {int res = 0;while(x) {res++;x /= 10;}return res;
}int main() {int n, p;while(~scanf("%d%d", &n, &p)) {int l = get(p);if(l > n) {printf("T_T\n");continue;}printf("%d", p);for(int i = l+1; i <= n; i++) {printf("0");}printf("\n");}return 0;
}

【E】 Find the median 线段树

参考：http://keyblog.cn/article-189.html

题目花里胡哨的，其实就是给定两个数组 L，R和一个空序列，对于第 \(i\) 对 LR 来说，把 \([L_i,R_i]\) 中的每个数加入序列，然后输出当前序列中的中位数。

主要是写起来麻烦，一些 +1/-1 会很容易错，尤其是我这种使用 \(vector\) 离散化的，我是多插了一个 -1 让下标从 1 开始，这样或者用数组可以规避很多 +1/-1。具体实现看代码。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;const int maxn = 8e5+5;ll x[maxn], y[maxn];
ll val[maxn<<2], lazy[maxn<<2];
vector<ll> v;int getid(ll x) {return lower_bound(v.begin(), v.end(), x) - v.begin();
}void pushup(int rt) {val[rt] = val[rt<<1] + val[rt<<1|1];
}void pushdown(int l, int r, int rt) {int mid = (l+r) >> 1;val[rt<<1] += (v[mid]-v[l])*lazy[rt];val[rt<<1|1] += (v[r]-v[mid])*lazy[rt];lazy[rt<<1] += lazy[rt];lazy[rt<<1|1] += lazy[rt];lazy[rt] = 0;
}void build(int l, int r, int rt) {if(l == r) {val[rt] = lazy[rt] = 0;return ;}int mid = (l+r) >> 1;build(l, mid, rt<<1);build(mid+1, r, rt<<1|1);pushup(rt);
}void update(int L, int R, int l, int r, int rt) {if(L <= l && r-1 <= R) {val[rt] += (v[r]-v[l]);lazy[rt] ++;return ;}if(lazy[rt]) {pushdown(l, r, rt);}int mid = (l+r) >> 1;if(L < mid) {update(L, R, l, mid, rt<<1);}if(R >= mid) {update(L, R, mid, r, rt<<1|1);}pushup(rt);
}ll query(int l, int r, int rt, ll x) {      // 找第 x 大的数if(l == r-1) {ll t = val[rt] / (v[r]-v[l]);return v[l] + (x-1)/t;}if(lazy[rt]) {pushdown(l, r, rt);}int mid = (l+r) >> 1;if(val[rt<<1] >= x) {return query(l, mid, rt<<1, x);}else {return query(mid, r, rt<<1|1, x-val[rt<<1]);}
}int main() {int n;scanf("%d", &n);v.clear();ll a1, b1, c1, m1;ll a2, b2, c2, m2;scanf("%lld%lld%lld%lld%lld%lld", &x[1], &x[2], &a1, &b1, &c1, &m1);scanf("%lld%lld%lld%lld%lld%lld", &y[1], &y[2], &a2, &b2, &c2, &m2);for(int i = 3; i <= n; i++) {x[i] = (a1*x[i-1]+b1*x[i-2]+c1) % m1;y[i] = (a2*y[i-1]+b2*y[i-2]+c2) % m2;}for(int i = 1; i <= n; i++) {ll tx = x[i];ll ty = y[i];x[i] = min(tx, ty) + 1;y[i] = max(tx, ty) + 1 + 1; // 多 +1 方便区间操作v.push_back(x[i]);v.push_back(y[i]);}v.push_back(-1);sort(v.begin(), v.end());v.erase(unique(v.begin(), v.end()), v.end());int new_n = v.size();ll sum = 0;build(1, new_n, 1);for(int i = 1; i <= n; i++) {int l = getid(x[i]);int r = getid(y[i]);update(l, r-1, 1, new_n, 1);sum = sum + (v[r] - v[l]);printf("%lld\n", query(1, new_n, 1, (sum+1)/2));}return 0;
}

【J】 A+B problem 水题

其实不是很想把它叫做水题，毕竟把我给卡了，明明 10 行搞定的东西，脑子一抽写了快 100 行，多花了不知道多久...以下代码为赛后补题

水题不写了。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;ll f(ll x) {ll res = 0;while(x) {res = res*10 + x%10;x /= 10;}return res;
}int main() {ll a, b;int t;scanf("%d", &t);while(t--) {scanf("%lld%lld", &a, &b);printf("%lld\n", f(f(a)+f(b)));}return 0;
}