Codeforces Round #451 (Div. 2)

C. Phone Numbers (模拟+STL)
思路：
一种节省空间的做法是map<string, int> + set<string> s[maxn]
筛选时，对每一个只需判断是否为后缀即可，判后缀可用string.substr()再进行比较.

#include <set>#include <map>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <deque>#include <cstdio>#include <string>#include <vector>#include <cctype>#include <sstream>#include <utility>#include <cstring>#include <cstdlib>#include <functional>#include <iostream>#include <algorithm>#define SF(a) scanf("%d", &a)#define PF(a) printf("%d\n", a)  #define SFF(a, b) scanf("%d%d", &a, &b)  #define SFFF(a, b, c) scanf("%d%d%d", &a, &b, &c)#define SFFFF(a, b, c, d) scanf("%d%d%d%d", &a, &b, &c, &d)#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define FOR(i, a, b) for(int i = a; i < b; ++i)#define ALL(a) (a).begin(), (a).end()#define SZ(a) (int)(a).size()#define PB push_back#define LL long long#define mod 10007#define inf 100000007#define eps 1e-12using namespace std;int buf[20];int read() {int x = 0; char ch = getchar(); bool f = 0;while (ch < '0' || ch > '9') { if (ch == '-') f = 1; ch = getchar(); }while (ch >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();return f ? -x : x;}void write(int x) {if (!x) { putchar(48); return; }int l = 0; if (x < 0) putchar('-'), x = -x;while (x) buf[++l] = x % 10, x = x / 10;while (l) putchar(buf[l--] + 48);}//-------------------------chc------------------------------//map<string, set<string>> M;int main() {int n;while (cin >> n) {while (n--) {string name;cin >> name;int m;cin >> m;while (m--) {string s;cin >> s;M[name].insert(s);}}cout << M.size() << endl;for (auto &it : M) {vector<string> v;for(string a : it.second) {bool ok = true;for(string b : it.second) {if (a == b) continue;if (SZ(b) > SZ(a) && b.substr(SZ(b) - SZ(a)) == a) {ok = false;break;}}if (ok) v.PB(a);}cout << it.first << ' ';cout << SZ(v);FOR(i, 0, v.size())cout << ' ' << v[i];cout << endl;}}return 0;}

B. Proper Nutrition (扩展欧几里得)
欧几里得算法: 
设a = p * b + q;
gcd(b, q) | a, gcd(b, q) | b -> gcd(b, q) | gcd(a, b);
反之q = a - p * b;
gcd(a, b) | q, gcd(a, b) | b -> gcd(a, b) | gcd(b, q);
即gcd(a, b) = gcd(b, a%b);


扩展欧几里得算法: 
求解ax + by = gcd(a, b)
bx' + (a%b)y' = gcd(a, b)
ay' + b(x' - (a/b)*y') = gcd(a, b)

递归边界: b = 0时, x = 1, y = 0

#include <set>#include <map>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <deque>#include <cstdio>#include <string>#include <vector>#include <cctype>#include <sstream>#include <utility>#include <cstring>#include <cstdlib>#include <functional>#include <iostream>#include <algorithm>#define SF(a) scanf("%d", &a)#define PF(a) printf("%d\n", a)  #define SFF(a, b) scanf("%d%d", &a, &b)  #define SFFF(a, b, c) scanf("%d%d%d", &a, &b, &c)#define SFFFF(a, b, c, d) scanf("%d%d%d%d", &a, &b, &c, &d)#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define FOR(i, a, b) for(int i = a; i < b; ++i)#define ALL(a) (a).begin(), (a).end()#define SZ(a) (int)(a).size()#define PB push_back#define LL long long#define mod 10007#define inf 100000007#define eps 1e-12using namespace std;int buf[20];int read() {int x = 0; char ch = getchar(); bool f = 0;while (ch < '0' || ch > '9') { if (ch == '-') f = 1; ch = getchar(); }while (ch >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();return f ? -x : x;}void write(int x) {if (!x) { putchar(48); return; }int l = 0; if (x < 0) putchar('-'), x = -x;while (x) buf[++l] = x % 10, x = x / 10;while (l) putchar(buf[l--] + 48);}//-------------------------chc------------------------------//void exgcd(LL a, LL b, LL &d, LL &x, LL &y) {if (b == 0) d = a, x = 1, y = 0;else {exgcd(b, a % b, d, y, x);y -= a / b*x;}}int main() {LL n, a, b, d, x, y;while (cin >> n >> a >> b) {exgcd(a, b, d, x, y);if (n % d) {puts("NO");continue;}x *= n / d, y *= n / d;if (x < 0) {while (x < 0) x += b, y -= a;}else if (y < 0) {while (y < 0) y += a, x -= b;}if (x >= 0 && y >= 0) {puts("YES");cout << x << ' ' << y << endl;}else puts("NO");}return 0;}  

其实很水。。。暴力就行。。

#include <set>#include <map>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <deque>#include <cstdio>#include <string>#include <vector>#include <cctype>#include <sstream>#include <utility>#include <cstring>#include <cstdlib>#include <functional>#include <iostream>#include <algorithm>#define SF(a) scanf("%d", &a)#define PF(a) printf("%d\n", a)  #define SFF(a, b) scanf("%d%d", &a, &b)  #define SFFF(a, b, c) scanf("%d%d%d", &a, &b, &c)#define SFFFF(a, b, c, d) scanf("%d%d%d%d", &a, &b, &c, &d)#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define FOR(i, a, b) for(int i = a; i < b; ++i)#define ALL(a) (a).begin(), (a).end()#define SZ(a) (int)(a).size()#define PB push_back#define LL long long#define mod 10007#define inf 100000007#define eps 1e-12using namespace std;int buf[20];int read() {int x = 0; char ch = getchar(); bool f = 0;while (ch < '0' || ch > '9') { if (ch == '-') f = 1; ch = getchar(); }while (ch >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();return f ? -x : x;}void write(int x) {if (!x) { putchar(48); return; }int l = 0; if (x < 0) putchar('-'), x = -x;while (x) buf[++l] = x % 10, x = x / 10;while (l) putchar(buf[l--] + 48);}//-------------------------chc------------------------------//void exgcd(LL a, LL b, LL &d, LL &x, LL &y) {if (b == 0) d = a, x = 1, y = 0;else {exgcd(b, a % b, d, y, x);y -= a / b*x;}}int main() {LL n, a, b;cin >> n >> a >> b;for (int i = 0; i * a <= n; ++i) {if ((n - i*a) % b == 0) {puts("YES");cout << i << ' ' << (n - i*a) / b << endl;return 0;}}puts("NO");return 0;}