Solutaion

A. Creating a Character

題意：
給出初始體力值\(str\)和智力值\(int\)，然后你可以把\(exp\)分別分配給這兩個數值，使得分配后\(str > int\)，求有多少種分配方案。
思路：

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

正解：
假設分別分配給\(str,int\)的數值為\(Adds,Addi\)，那么有
\[ \begin{align*} & str + Adds > int + Addi \\ {\Rightarrow}{\quad} & str + Adds > int + (exp - Adds)\\ {\Rightarrow}{\quad} &2{\ast}Adds > int + exp - str\\ {\Rightarrow}{\quad} &2{\ast}Adds\ {\geq}\ int + exp - str + 1\\ {\Rightarrow}{\quad} &Adds\ {\geq}\ {\lceil}{\frac{int + exp - str + 1}{2}}{\rceil}\\ {\Rightarrow}{\quad} &Adds\ {\geq}\ {\frac{int + exp - str + 1 + 1}{2}} \end{align*} \]
因為非負，所以\(Adds=max(0,{\frac{int + exp - str + 2}{2}})\)，定義這個值為\(minAdds\)，分配值的區間為\([minAdds,exp]\)，那么答案為\(ans=max(0,exp - minAdds + 1)\)。

//#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

題意：
讓你斬殺多頭蛇，給出頭數\(x\)和你可以斬殺的類型\(n\)。每種類型包含兩個數\(d,h\)，代表每次斬殺能斬掉\(d\)個頭，如果沒死的話，他會長出\(h\)個頭。問最少斬殺次數。
思路：
首先，如果第一次用最大“毛斬殺”可以殺死就結束了，如果不能殺死，就用最大“純斬殺”來斬。

//#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

題意：
假定\(f(t)=val\)，其中\(t\)為\(01\)字符串，\(val\)為其代表的二進制值，比如\(f(011)=3，f(00101) = 5\)。給出一個\(01\)字符串，求有多少個子串使得\(f(s_l,s_{l+1},{\dots},s_r) = r - l + 1\)
思路：
因為字符串長度不超過\(2e5\)，所以可以每次枚舉20位去判斷。預處理出\(nxt[i]\)，表示\(1{\dots}i\)中最后一個\(1\)的位置，\(nxt[i]=-1\)。枚舉\(i\)前\(20\)位，定義\(sum\)為當前長度子串所代表的二進制的值。如果當前\(sum<=r-nxt[l]\)，貢獻\(+1\)。

//#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

題意：
給出\(n\)個點\(m\)條邊的有向圖，然后給邊染色，用最少種類的顏料，使得環上的邊不是純色。求最少種類。
思路：
畫出圖可以分析出，不存在環顯然一種即可。若存在環，最多需要兩種顏料。在發現環的時候換色即可。好像之前做過類似的題目。但是比賽時沒有看這個題。\(dfs\)先一次標記該點，用顏料1一直染邊，如果遇到某點被一次標記，說明存在環，該邊染為顏料2。如果遇到二次標記點，該邊染為顏料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?

題意：
如果集合內元素右對齊放置，對應位出現的數字之和不等于對應位的非\(0\)數字，則說明該可重復元素集合是\(unbalanced\)。換句話說，就是如果放置后，對應位的數字有兩個非\(0\)數字，就說明\(unbalanced\)。可單點修改某一位置的值，求每次詢問區間的不平衡集合的最小和。
思路：
就是區間找兩個對應位都不為\(0\)的數字，然后求最小和。因為\(a_i{\leq}10^9\)網上都是說開\(10\)棵線段樹，我不是很理解這個說法，把數字\(x\)拆分開。如果某位數字不為0，就設為\(x\)，否則設為\(inf\)，然后維護每一位的最小值，維護答案。初始為\(inf\)，單點更新和建樹差不多。\(pushup\)操作就\(Min[rt][i]\)為左右兒子的對應的最小值，\(val[rt]\)就是左右兒子對應為都不為\(inf\)時的和，同時也是左右兒子的答案的最小值。\(query\)操作用\(res[i]\)保存這次查詢歷史對應位的最小值，然后\(ans=min(ans,res[i]+Min[rt][i])\)，每次再更新\(res[i]\)。用我的\(inf\)會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);}} }

轉載于:https://www.cnblogs.com/ACMerszl/p/11517315.html

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