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Educational Codeforces Round 72 (Rated for Div. 2)

时间:2019-09-13 19:31:40      阅读:77      评论:0      收藏:0      [点我收藏+]

标签:scanf   rac   标记   tmp   tar   cos   efi   出现   c++   

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 DEBUG
    freopen("in.txt", "r", stdin);
    #endif
    for (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 DEBUG
    freopen("in.txt", "r", stdin);
    #endif
    for (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 DEBUG
    freopen("in.txt", "r", stdin);
    #endif
    for (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\(t\)\(01\)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\(i\)\(20\)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 DEBUG
    freopen("in.txt", "r", stdin);
    #endif
    for (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 DEBUG
    freopen("in.txt", "r", stdin);
    #endif
    flag = 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 DEBUG
    freopen("in.txt", "r", stdin);
    #endif
    scanf("%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);
        }
    }   
}

Educational Codeforces Round 72 (Rated for Div. 2)

标签:scanf   rac   标记   tmp   tar   cos   efi   出现   c++   

原文地址:https://www.cnblogs.com/ACMerszl/p/11517315.html

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