CCF CSP第十一次认证考试 20170917

第一题： 买酱油，买五瓶送两瓶，买三瓶送一瓶。10元/瓶。给出钱数，问最多能买多少瓶酱油。

第二题：教师取放钥匙，结构体排序，重载运算符。

第三题：文件操作，字符串处理，查询。

第四题：有向图的搜索。能达到点v的点数 + 点v可达到的点数 == 总点数n - 1。询问有多少个这样的点。暴力深搜每个点，记录可达信息

第五题：听说特判1能得八十分。用树状数组+特判1优化就满分了。好像还可以优雅地暴力解题，分块+莫队。但我不会。

第三题把读到的信息去掉所有空格，就是一个串。对于要查询的字符串例如address，变成 键"address": 这样就不会和值中的address重了。只得了60分这题。 

第一题：（100）

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <list>#include <cmath>using namespace std;int main(){//ios::sync_with_stdio(false);int n;while(~scanf("%d", &n)) {n /= 10;int cnt = (n / 5) * 7;n %= 5;cnt += (n / 3) * 4;n %= 3;cnt += n;printf("%d\n", cnt);}return 0;}

第二题：（100）

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <list>#include <cmath>using namespace std;const int N = 1007;struct Node {int id, t, fg; //t时间点， fg: 0还钥匙，1取钥匙 Node(int id = 0, int t = 0, int fg = 0):id(id), t(t), fg(fg){}bool operator < (const Node &tmp) const {return t == tmp.t ? (fg == tmp.fg ? id < tmp.id : fg < tmp.fg) : t < tmp.t;}}a[N << 1];int v[N];int main(){//ios::sync_with_stdio(false);int n, k;while(~scanf("%d%d", &n, &k)) {int id, st, len, cur = 0;for(int i = 1; i <= k; ++i) {scanf("%d%d%d", &id, &st, &len);a[cur++] = Node(id, st, 1);a[cur++] = Node(id, st + len, 0);}for (int i = 1; i <= n; ++i) v[i] = i;sort(a, a + cur);//for (int i = 0; i < cur; ++i) {//printf("%d %d %d\n", a[i].t, a[i].id, a[i].fg);//}for (int i = 0; i < cur; ++i) {if(a[i].fg == 1) {for (int pos = 1; pos <= n; ++pos) if(v[pos] == a[i].id) { v[pos] = 0;  break;}}else {for (int pos = 1; pos <= n; ++pos) if(v[pos] == 0) {v[pos] = a[i].id;break;}}}for(int i = 1; i <= n; ++i) {printf("%d%c", v[i], " \n"[i==n]);}}return 0;}

第三题：（60）

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <list>#include <cmath>using namespace std;const int N = 107;string dup(string str) {for (string::iterator it = str.begin(); it != str.end(); ++it) {if(*it == ' ') str.erase(it--);}return str;}string add(string str) {string res = "";int len = str.length();for (int i = 0; i < str.length(); ++i) {if(str[i] == '\\' || str[i] == '\"') res += '\\';res += str[i];}return res;}int main(){ios::sync_with_stdio(false);int n, m;while(cin >> n >> m) {string tmp, s = "", t;cin.get();for (int i = 0; i < n; ++i) {getline(cin, tmp); s += tmp;}s = dup(s); //cout << s << endl << s.length() << endl;int len = s.length();while(m-- > 0) {cin >> t; t = add(t);///*t = '\"' + dup(t) + "\":";// cout << t << endl;string res = "";int found = (int) s.find(t), i;if(found == -1) {cout << "NOTEXIST" << endl;}else {found += t.length();//cout << found << " " << s[found] << endl;for (i = found; i < len; ++i) {if(s[i] == '}' && i != len - 1) {found = (int)s.find(t, i);//cout << i << " " << s[i] << endl;if(found == -1) {cout << "NOTEXIST" << endl; }else {found += t.length() + 1;if(s[found] == '{') cout << "OBJECT" << endl;else {   cout << "STRING ";while(s[found] != '\"') cout << s[found++];cout << endl;//error}} break;}else if(s[i] == '{'){break;}}if(i >= len || s[i] == '{') {//cout << found << " " << s[found] << endl;if(s[found] == '{') cout << "OBJECT" << endl;else {cout << "STRING ";for(int i= found + 1; i < len; ++i) {if(s[i] == '\\') cout << s[++i];else if(s[i] == '\"') break;else cout << s[i];}cout << endl;}} //else {//cout << i << " " << s[i] << endl;//}}//*/}}return 0;}

第四题：（100）

#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <vector>#include <cstring>using namespace std;const int N = 1007;vector<int> e[N]; //动态数组邻接表存图bool mp[N][N]; //mp[i][j] i 到 j是否通 bool vis[N];void dfs(int u, int cur) {vis[u] = 1;mp[u][cur] = mp[cur][u] = 1;for (int i = e[u].size() - 1; i >= 0; --i) {int v = e[u][i];if (!vis[v])dfs(v, cur);}}int main(){int n, m, cnt = 0;scanf("%d%d", &n, &m);for (int i = 0; i<m; ++i) {int u, v;scanf("%d%d", &u, &v);e[u].push_back(v);}for (int i = 1; i <= n; ++i) {memset(vis, 0, sizeof vis);dfs(i, i);}for (int i = 1; i <= n; ++i) {bool fg = true;;for (int j = 1; j <= n; ++j) {if (mp[i][j] == 0) {fg = false;break;}}cnt += fg;}printf("%d", cnt);return 0;}

第五题：

#include <cstdio>#include <cstring>using namespace std;const int N = 1e6 + 7;typedef long long LL;LL bit[N];int n, m, a[N];inline int lowbit(int x) {    return x & -x;}void add(int pos, int val) {    for (int i = pos; i <= n; i += lowbit(i))        bit[i] += val;}LL sum(int x) {    LL res = 0;    for (int i = x; i > 0; i -= lowbit(i)) {        res += bit[i];    }    return res;}int main(){    while(~scanf("%d%d", &n, &m)) {        memset(bit, 0, sizeof bit);        for (int i = 1; i <= n; ++i) {            scanf("%d", a + i);            add(i, a[i]);        }        while(m-- > 0) {            int opt, l, r, val;            scanf("%d", &opt);            if(opt == 1) {                scanf("%d%d%d", &l, &r, &val);                if(val == 1) continue; //虽有整数都是1的倍数，优化                for(int i = l; i <= r; ++i) if(a[i] >= val && a[i] % val == 0) {                        add(i,a[i] / val - a[i]);                        a[i] /= val;                }            }            else if(opt == 2) {                scanf("%d%d", &l, &r);                printf("%lld\n", sum(r) - sum(l - 1));            }        }    }    return 0;}