Codeforces Round #360 div2
Problem_A(CodeForces 688A):
题意:
有d天, n个人。如果这n个人同时出现, 那么你就赢不了他们所有的人, 除此之外, 你可以赢他们所有到场的人。
到场人数为0也算赢。
现给出这n个人d天的到勤情况, 求最大连胜天数。
思路:
暴力找下去, 维护最大天数即可。
代码:
#include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <ctime> #include <set> #include <map> #include <list> #include <stack> #include <queue> #include <string> #include <vector> #include <fstream> #include <iterator> #include <iostream> #include <algorithm> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define MOD 1000000007 #define eps 1e-6 #define MAXN 110 #define MAXM 100 #define dd {cout<<"debug"<<endl;} #define pa {system("pause");} #define p(x) {printf("%d\n", x);} #define pd(x) {printf("%.7lf\n", x);} #define k(x) {printf("Case %d: ", ++x);} #define s(x) {scanf("%d", &x);} #define sd(x) {scanf("%lf", &x);} #define mes(x, d) {memset(x, d, sizeof(x));} #define do(i, x) for(i = 0; i < x; i ++) #define dod(i, x, l) for(i = x; i >= l; i --) #define doe(i, x) for(i = 1; i <= x; i ++) int n, d; int main() { char str[MAXN]; int ans = 0, max_day = 0; scanf("%d %d", &n, &d); for(int i = 0; i < d; i ++) { scanf("%s", str); bool flag = false; for(int j = 0; j < n; j ++) if(str[j] == '0') flag = true; if(!flag) { ans = 0; } else ans = ans + 1; max_day = max(ans, max_day); } printf("%d\n", max_day); return 0; }
Problem_B(CodeForces 688B):
题意:
给你一个n, 给出第n个偶数长度回文串。
思路:
显而易见, 第n个回文串就是n+n的反转, 反向再输出一次即可。
代码:
#include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <ctime> #include <set> #include <map> #include <list> #include <stack> #include <queue> #include <string> #include <vector> #include <fstream> #include <iterator> #include <iostream> #include <algorithm> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define MOD 1000000007 #define eps 1e-6 #define MAXN 1000010 #define MAXM 100 #define dd {cout<<"debug"<<endl;} #define pa {system("pause");} #define p(x) {printf("%d\n", x);} #define pd(x) {printf("%.7lf\n", x);} #define k(x) {printf("Case %d: ", ++x);} #define s(x) {scanf("%d", &x);} #define sd(x) {scanf("%lf", &x);} #define mes(x, d) {memset(x, d, sizeof(x));} #define do(i, x) for(i = 0; i < x; i ++) #define dod(i, x, l) for(i = x; i >= l; i --) #define doe(i, x) for(i = 1; i <= x; i ++) int len; char str[MAXN]; int main() { scanf("%s", str); printf("%s", str); for(int i = strlen(str) - 1; i >= 0; i --) printf("%c", str[i]); printf("\n"); return 0; }
Problem_C(CodeForces 688C):
题意:
给一个图, n个点,m条边。
要求你找到这样的两个集合 A, B。
每个集合都满足如下条件:
任意一条边至少有一个端点在这个集合中。
并且A, B无交集。
思路:
种类并查集, 先将其分成两个类。
然后对于每条边, 看它们是否在同一个类里, 如果在同一个类里, 那么就不可能找到这样的两个集合(因为A, B都要满足上述条件)。
不在同一个集合便分别加入两个类里。
代码:
#include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <ctime> #include <set> #include <map> #include <list> #include <stack> #include <queue> #include <string> #include <vector> #include <fstream> #include <iterator> #include <iostream> #include <algorithm> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define MOD 1000000007 #define eps 1e-6 #define MAXN 400010 #define MAXM 100 int n, m; int fa[2 * MAXN]; bool has[MAXN]; int A[MAXN], B[MAXN]; int cnt_a, cnt_b; int find_(int x) { return fa[x] = x == fa[x] ? fa[x] : find_(fa[x]); } void union_(int x, int y) { x = find_(x); y = find_(y); if(x != y) fa[y] = x; } bool same(int x, int y) { return find_(x) == find_(y); } int main() { memset(has, false, sizeof(has)); cnt_a = 0, cnt_b = 0; scanf("%d %d", &n, &m); for(int i = 0; i < 2 * MAXN; i ++) fa[i] = i; int u, v; scanf("%d %d", &u, &v); union_(u, v + n); union_(u + n, v); has[u] = has[v] = true; bool flag = false; for(int i = 1; i < m; i ++) { scanf("%d %d", &u, &v); if(flag) continue; if(same(u, v)) flag = true; else { union_(u, v + n); union_(u + n, v); has[u] = has[v] = true; } } if(flag) printf("-1\n"); else { for(int i = 1; i <= n; i ++) { if(has[i] && find_(i) <= n) A[cnt_a ++] = i; if(has[i] && find_(i) > n) B[cnt_b ++] = i; } printf("%d\n", cnt_a); for(int i = 0; i < cnt_a; i ++) printf("%d ", A[i]); printf("\n%d\n", cnt_b); for(int i = 0; i < cnt_b; i ++) printf("%d ", B[i]); printf("\n"); } return 0; }
Problem_D(CodeForces 688D):
题意:
给n个ci, 可以假设已知 x % ci = ai。
现给一个k, 问能否由这n个式子确定x % k的值。
思路:
\(由题意可知,如果存在这样的x_1\space x_2\)
\(使得\forall _{i\in [1,n]} 有 x_1\equiv a_i(mod\space c_i) 且x_2\equiv a_i(mod \space c_i)\)
$\because $
\(\left\{ \begin{array}{c} x_1\equiv a_i(mod \space c_i)\\ x_2\equiv a_i(mod \space c_i)\\ \end{array} \right.\)
得如下式子:
\(\left\{ \begin{array}{c} x_1 \%c_i=a_i\%c_i\\ x_2 \%c_i=a_i\%c_i\\ \end{array} 令b=a_i\%c_i得\longrightarrow \{ \begin{array}{c} x_1\%c_i=b\\ x_2\%c_i=b\\ \end{array} \right.\)
\(\therefore (x_1 -x_2) \equiv 0(mod \space c_i)\)
\(由此可得, (x_1-x_2)=yc_i \longrightarrow c_i \mid (x_1-x_2)\)
\(\because \forall _{i\in [1, n]} 都有c_i \mid (x_1-x_2) \longrightarrow lcm(c_1,c_2,\cdots,c_n)\mid(x_1-x_2)\)
\(如果有\)
\(x_1\equiv b(mod \space k)\)
\(x_2\equiv c(mod \space k)\)
\(b\neq c时,即由这n个c_i不能确定x\%k的值\)
\(即(x_1-x_2) \neq0(mod \space k) \longrightarrow lcm(c_1,c_2,\cdots, c_n)\nmid k\)
\(b=c时,表示可以确定x\%k的值\)
\(即lcm(c_1, c_2,\cdots, c_n)\mid k \longrightarrow lcm(c_i) \mid k = 0\)
数比较大, 所以需要边除边算
代码:
#include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <ctime> #include <set> #include <map> #include <list> #include <stack> #include <queue> #include <string> #include <vector> #include <fstream> #include <iterator> #include <iostream> #include <algorithm> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define MOD 1000000007 #define eps 1e-6 #define MAXN 1000000 #define MAXM 100 #define dd {cout<<"debug"<<endl;} #define pa {system("pause");} #define p(x) {printf("%d\n", x);} #define pd(x) {printf("%.7lf\n", x);} #define k(x) {printf("Case %d: ", ++x);} #define s(x) {scanf("%d", &x);} #define sd(x) {scanf("%lf", &x);} #define mes(x, d) {memset(x, d, sizeof(x));} #define do(i, x) for(i = 0; i < x; i ++) #define dod(i, x, l) for(i = x; i >= l; i --) #define doe(i, x) for(i = 1; i <= x; i ++) int n, k; LL gcd(LL a, LL b) { return b == 0? a : gcd(b, a % b); } LL lcm(LL a, LL b) { return a / gcd(a, b) * b; } int main() { scanf("%d %d", &n, &k); int ans = 1; int c; for(int i = 0; i < n; i ++) { scanf("%d", &c); ans = gcd(k, lcm(ans, c)); } printf(ans == k? "Yes\n" : "No\n"); return 0; }
Problem_E(CodeForces 688E):
题意:
给n个硬币,让你用这n个硬币组合出k。
并且对于每个能组合出k的组合, 计算出它能够组合出来的所有数。
思路:
设dp[i][j][y]为从前1~i个硬币, 和为sum时, 能否组合出y。
那么dp[i][j][y]就由三个状态转移过来。
1、不选第i个硬币(dp[i-1][j][y])
2、选择第i个硬币,但是集合中已经有c[i]了(dp[i-1][j-c[i]][y])
3、选择第i个硬币,集合中不存在ci
代码:
#include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <ctime> #include <set> #include <map> #include <list> #include <stack> #include <queue> #include <string> #include <vector> #include <fstream> #include <iterator> #include <iostream> #include <algorithm> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define MOD 1000000007 #define eps 1e-6 #define MAXN 510 #define MAXM 100 #define dd {cout<<"debug"<<endl;} #define pa {system("pause");} #define p(x) {printf("%d\n", x);} #define pd(x) {printf("%.7lf\n", x);} #define k(x) {printf("Case %d: ", ++x);} #define s(x) {scanf("%d", &x);} #define sd(x) {scanf("%lf", &x);} #define mes(x, d) {memset(x, d, sizeof(x));} #define do(i, x) for(i = 0; i < x; i ++) #define dod(i, x, l) for(i = x; i >= l; i --) #define doe(i, x) for(i = 1; i <= x; i ++) int n, k; bool dp[2][MAXN][MAXN]; int main() { scanf("%d %d", &n, &k); int c; dp[0][0][0] = 1; for(int i = 1; i <= n; i ++) { int cnt = i % 2; int pre = 1 - cnt; scanf("%d", &c); for(int j = 0; j <= k; j ++) for(int y = 0; y <= j; y ++) { dp[cnt][j][y] = dp[pre][j][y]; if(j >= c) dp[cnt][j][y] = (dp[cnt][j][y] | dp[pre][j - c][y]) | (y >= c? dp[pre][j - c][y - c] : 0); } } vector <int> V; for(int i = 0; i <= k; i ++) if(dp[n % 2][k][i]) V.push_back(i); printf("%d\n", V.size()); for(int i = 0; i < V.size(); i ++) printf("%d ", V[i]); printf("\n"); return 0; }
- Codeforces Round #360 (Div. 2) D. Remainders Game 数学
- Codeforces Round #360 (Div. 2)
- Codeforces Round #360 (Div. 2)-B. Lovely Palindromes
- Codeforces Round #360 (Div. 2) -- D. Remainders Game (中国剩余定理)
- Codeforces Round #360 (Div. 1)A (二分图&dfs染色)
- Codeforces Round #360 (Div. 2) E The Values You Can Make(DP)
- Codeforces Round #360(div2)
- Codeforces Round #360 (Div. 2) E. The Values You Can Make DP
- Codeforces Round #360 (Div. 2) E. The Values You Can Make 01背包
- Codeforces Round #360 (Div. 1)
- Codeforces Round #360 (Div. 2) D. Remainders Game(数学)
- Codeforces Round #360 (Div. 1) 题解(待续)
- Codeforces Round #360 (Div. 2) E. The Values You Can Make dp
- Codeforces Round #360 (Div. 1) A. NP-Hard Problem(二分图染色)
- Codeforces Round #360 (Div. 2) A. Opponents(水题)
- Codeforces Round #360 (Div. 2) -- C. NP-Hard Problem (DFS二分图染色法)
- Codeforces Round #360 (Div. 1) D 并查集判奇环
- Codeforces Round #360 (Div. 1) D. Dividing Kingdom II 暴力,二分图,并查集
- Codeforces Round #360 (Div. 2) E. The Values You Can Make dp ,滚动数组
- Codeforces Round #360 (Div. 2) B Lovely Palindromes