Solutaion

A. Creating a Character

题意：
给出初始体力值\(str\)和智力值\(int\)，然后你可以把\(exp\)分别分配给这两个数值，使得分配后\(str > int\)，求有多少种分配方案。
思路：

• 特判不可能情况：\(str + exp <= int\)
• \(str <= int\)，乱搞
• \(str > int\)，乱搞

正解：
假设分别分配给\(str,int\)的数值为\(Adds,Addi\)，那么有
\[ \begin{align*} & str + Adds > int + Addi \\ {\Rightarrow}{\quad} & str + Adds > int + (exp - Adds)\\ {\Rightarrow}{\quad} &2{\ast}Adds > int + exp - str\\ {\Rightarrow}{\quad} &2{\ast}Adds\ {\geq}\ int + exp - str + 1\\ {\Rightarrow}{\quad} &Adds\ {\geq}\ {\lceil}{\frac{int + exp - str + 1}{2}}{\rceil}\\ {\Rightarrow}{\quad} &Adds\ {\geq}\ {\frac{int + exp - str + 1 + 1}{2}} \end{align*} \]
因为非负，所以\(Adds=max(0,{\frac{int + exp - str + 2}{2}})\)，定义这个值为\(minAdds\)，分配值的区间为\([minAdds,exp]\)，那么答案为\(ans=max(0,exp - minAdds + 1)\)。

//#define DEBUG
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = 0X3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1000000007;
typedef long long ll;int _;
int main() {#ifdef DEBUGfreopen("in.txt", "r", stdin);#endiffor (scanf("%d", &_); _; _--) {ll str, intt, exp;scanf("%lld %lld %lld", &str, &intt, &exp);if (str + exp <= intt) puts("0");else if (str <= intt) {ll x =  exp - (intt - str);printf("%lld\n", x % 2 ? x / 2 + 1 : x / 2);} else {if (intt + exp - str < 0) printf("%lld\n", exp + 1);else {ll x = (intt + exp - str) / 2 + 1;printf("%lld\n", exp - x + 1);}}}
}

//#define DEBUG
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = 0X3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1000000007;
typedef long long ll;int _;
int main() {#ifdef DEBUGfreopen("in.txt", "r", stdin);#endiffor (scanf("%d", &_); _; _--) {int st, in, ex, tmp;scanf("%d %d %d", &st, &in, &ex);tmp = max(0, (in + ex - st + 2) / 2);printf("%d\n", max(ex - tmp + 1, 0));}
}

B. Zmei Gorynich

题意：
让你斩杀多头蛇，给出头数\(x\)和你可以斩杀的类型\(n\)。每种类型包含两个数\(d,h\)，代表每次斩杀能斩掉\(d\)个头，如果没死的话，他会长出\(h\)个头。问最少斩杀次数。
思路：
首先，如果第一次用最大“毛斩杀”可以杀死就结束了，如果不能杀死，就用最大“纯斩杀”来斩。

//#define DEBUG
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = 0X3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1000000007;
typedef long long ll;int _;
int main() {#ifdef DEBUGfreopen("in.txt", "r", stdin);#endiffor (scanf("%d", &_); _; _--) {int m, n;scanf("%d %d", &m, &n);int val = -inf, maxx = -inf;while (m--) {int u, v;scanf("%d %d", &u, &v);val = max(val, u - v);maxx = max(maxx, u);}ll ans = 1;n -= maxx;if (n > 0) {if (val <= 0) ans = -1;else ans += (n + val - 1) / val;}printf("%lld\n", ans);}
}

C. The Number Of Good Substrings

题意：
假定\(f(t)=val\)，其中\(t\)为\(01\)字符串，\(val\)为其代表的二进制值，比如\(f(011)=3，f(00101) = 5\)。给出一个\(01\)字符串，求有多少个子串使得\(f(s_l,s_{l+1},{\dots},s_r) = r - l + 1\)
思路：
因为字符串长度不超过\(2e5\)，所以可以每次枚举20位去判断。预处理出\(nxt[i]\)，表示\(1{\dots}i\)中最后一个\(1\)的位置，\(nxt[i]=-1\)。枚举\(i\)前\(20\)位，定义\(sum\)为当前长度子串所代表的二进制的值。如果当前\(sum<=r-nxt[l]\)，贡献\(+1\)。

//#define DEBUG
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = 0X3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1000000007;
typedef long long ll;int _;
char s[2 * N];
int nxt[2 * N];int main() {#ifdef DEBUGfreopen("in.txt", "r", stdin);#endiffor (scanf("%d", &_); _; _--) {scanf("%s", s);int len = strlen(s);for (int i = 0; i < len; i++) {if (s[i] == '0') nxt[i] = (i == 0 ? -1 : nxt[i - 1]);else nxt[i] = i;}int ans = 0;for (int i = 0; i < len; i++) {int sum = 0;for (int j = i; j >= 0 && i - j + 1 <= 20; j--) {if (s[j] == '0') continue;sum += 1 << (i - j);if (sum <= i - (j == 0 ? -1 : nxt[j - 1]))ans++;}}printf("%d\n", ans);}
}

D. Coloring Edges

题意：
给出\(n\)个点\(m\)条边的有向图，然后给边染色，用最少种类的颜料，使得环上的边不是纯色。求最少种类。
思路：
画出图可以分析出，不存在环显然一种即可。若存在环，最多需要两种颜料。在发现环的时候换色即可。好像之前做过类似的题目。但是比赛时没有看这个题。\(dfs\)先一次标记该点，用颜料1一直染边，如果遇到某点被一次标记，说明存在环，该边染为颜料2。如果遇到二次标记点，该边染为颜料1。

//#define DEBUG
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = 0X3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1000000007;
typedef long long ll;vector<pair<int, int> > G[5010];
int n, m;
int colour[5010];
int res[5010];
bool flag;void dfs(int u) {colour[u] = 1;for (auto it : G[u]) {int to = it.first, id = it.second;if (!colour[to]) {res[id] = 1;dfs(to);} else if (colour[to] == 1) {res[id] = 2;flag = true;} else {res[id] = 1;}}colour[u] = 2;
}int main() {#ifdef DEBUGfreopen("in.txt", "r", stdin);#endifflag = false;scanf("%d %d", &n, &m);for (int i = 1; i <= m; i++) {int u, v;scanf("%d %d", &u, &v);G[u].push_back(make_pair(v, i));}for (int i = 1; i <= n; i++) {if (!colour[i]) dfs(i);}if (flag) puts("2");else puts("1");for (int i = 1; i <= m; i++) printf("%d%c", res[i], i == m ? '\n' : ' ');
}

E. Sum Queries?

题意：
如果集合内元素右对齐放置，对应位出现的数字之和不等于对应位的非\(0\)数字，则说明该可重复元素集合是\(unbalanced\)。换句话说，就是如果放置后，对应位的数字有两个非\(0\)数字，就说明\(unbalanced\)。可单点修改某一位置的值，求每次询问区间的不平衡集合的最小和。
思路：
就是区间找两个对应位都不为\(0\)的数字，然后求最小和。因为\(a_i{\leq}10^9\)网上都是说开\(10\)棵线段树，我不是很理解这个说法，把数字\(x\)拆分开。如果某位数字不为0，就设为\(x\)，否则设为\(inf\)，然后维护每一位的最小值，维护答案。初始为\(inf\)，单点更新和建树差不多。\(pushup\)操作就\(Min[rt][i]\)为左右儿子的对应的最小值，\(val[rt]\)就是左右儿子对应为都不为\(inf\)时的和，同时也是左右儿子的答案的最小值。\(query\)操作用\(res[i]\)保存这次查询历史对应位的最小值，然后\(ans=min(ans,res[i]+Min[rt][i])\)，每次再更新\(res[i]\)。用我的\(inf\)会WA5，小于\(2e9\)。

//#define DEBUG
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const int inf = 0X3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1000000007;
typedef long long ll;ll val[2 * N * 4], Min[2 * N * 4][15];
int a[2 * N];
int n, m;
ll res[15];
ll ans;void pushup(int rt) {val[rt] = INF;for (int i = 0; i <= 12; i++) {if (Min[rt << 1][i] != INF && Min[rt << 1 | 1][i] != INF) {val[rt] = min(val[rt], Min[rt << 1][i] + Min[rt << 1 | 1][i]);}Min[rt][i] = min(Min[rt << 1][i], Min[rt << 1 | 1][i]);}val[rt] = min(val[rt], min(val[rt << 1], val[rt << 1 | 1]));
}void build(int l, int r, int rt) {val[rt] = INF;if (l == r) {int tmp = a[l];for (int i = 0; i <= 12; i++) {int x = tmp % 10;if (x == 0) Min[rt][i] = INF;else Min[rt][i] = a[l];tmp /= 10;}return ;}int mid = l + r >> 1;build(l, mid, rt << 1);build(mid + 1, r, rt << 1 | 1);pushup(rt);
}void modify(int pos, int x, int l, int r, int rt) {if (l == r) {int tmp = x;for (int i = 0; i <= 12; i++) {int y = tmp % 10;if (y == 0) Min[rt][i] = INF;else Min[rt][i] = x;tmp /= 10;}return ;}int mid = l + r >> 1;if (pos <= mid) modify(pos, x, l , mid, rt << 1);else modify(pos, x, mid + 1, r, rt << 1 | 1);pushup(rt);
}void query(int L, int R, int l, int r, int rt) { if (L <= l && r <= R) {for (int i = 0; i <= 12; i++) {if (Min[rt][i] != INF && res[i] != INF) {ans = min(ans, Min[rt][i] + res[i]);}}for (int i = 0; i <= 12; i++) {res[i] = min(res[i], Min[rt][i]);}ans = min(ans, val[rt]);return ;}int mid = l + r >> 1;if (L <= mid) query(L, R, l, mid, rt << 1);if (R > mid) query(L, R, mid + 1, r, rt << 1 | 1);
}int main() {#ifdef DEBUGfreopen("in.txt", "r", stdin);#endifscanf("%d %d", &n, &m);for (int i = 1; i <= n; i++) scanf("%d", &a[i]);build(1, n, 1);while (m--) {int op, l, r, pos, x;scanf("%d", &op);if (op == 1) {scanf("%d %d", &pos, &x);          modify(pos, x, 1, n, 1);} else {scanf("%d %d", &l, &r);ans = INF;for (int i = 0; i <= 12; i++) res[i] = INF;query(l, r, 1, n, 1);printf("%lld\n", ans == INF ? -1 : ans);}}
}