Educational Codeforces Round 29

A

把末尾的0删除然后匹配

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e4 + 10;const int INF = 1e9 + 10;char st[30];int main(){scanf("%s", st);int r = strlen(st) - 1;int l = 0;while(st[r] == '0')r--;bool f = true;while(l <= r) {if(st[l] != st[r])f = false;l++, r--;}if(f)puts("YES");elseputs("NO");return 0;}

B

枚举两个不参加的人

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e6 + 10;const int INF = 1e9 + 10;int num[60];int main(){int n;scanf("%d", &n);for(int i = 1; i <= 2 * n; ++i) {scanf("%d", num + i);}sort(num + 1, num + 1 + 2 * n);int minx = INF;for(int i = 1; i <= 2 * n; ++i) {for(int j = i + 1; j <= 2 * n; ++j) {int sum = 0, a, b;a = b = -1;for(int k = 1; k <= 2 * n; ++k) {if(k == i || k == j)continue;if(a == -1) {a = num[k];} else {b = num[k];sum += abs(a - b);a = b = -1;}}minx = min(minx, sum);}}printf("%d\n", minx);return 0;}

C

题意：k局比赛，第一局Alice出a，Bob出b，之后每一局两人出的数字仅仅只和前一局有关，给出相关矩阵，让你求k局后的最终分数

思路：很显然进行到某个程度之后肯定是有循环节的，所以我们进行比赛直到找到循环节为止，然后处理一下有不循环的部分和循环的部分就好了

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e6 + 10;const int INF = 1e9 + 10;LL k, a, b;int A[4][4], B[4][4];LL dp[10005][2];int Check(LL x, LL y) {if(x == 1 && y == 3)return 1;if(x == 2 && y == 1)return 1;if(x == 3 && y == 2)return 1;if(x == y)return 0;return 2;}map<pill, int> mp;int main(){scanf("%lld%lld%lld", &k, &a, &b);for(int i = 1; i <= 3; ++i) {for(int j = 1; j <= 3; ++j) {scanf("%d", &A[i][j]);}}for(int i = 1; i <= 3; ++i) {for(int j = 1; j <= 3; ++j) {scanf("%d", &B[i][j]);}}dp[0][0] = dp[0][1] = 0;int cnt = 1;while(true) {int t = Check(a, b);dp[cnt][0] = dp[cnt - 1][0];dp[cnt][1] = dp[cnt - 1][1];if(t == 1) {dp[cnt][0]++;} else if(t == 2) {dp[cnt][1]++;}if(mp.find(mk(a, b)) == mp.end()) {mp[mk(a, b)] = cnt;int x, y;x = A[a][b], y = B[a][b];a = x, b = y;cnt++;} else {break;}}LL x = dp[cnt - 1][0] - dp[mp[mk(a, b)] - 1][0];LL y = dp[cnt - 1][1] - dp[mp[mk(a, b)] - 1][1];int len = (cnt - 1) - mp[mk(a, b)] + 1;if(k <= cnt) {printf("%lld %lld\n", dp[k][0], dp[k][1]);} else {LL ans1, ans2;ans1 = dp[mp[mk(a, b)] - 1][0], ans2 = dp[mp[mk(a, b)] - 1][1];k -= mp[mk(a, b)] - 1;LL tmp = k / len;ans1 += x * tmp, ans2 += y * tmp;k = k - tmp * len;for(int i = mp[mk(a, b)]; k > 0; ++i, --k) {ans1 += dp[i][0] - dp[i - 1][0];ans2 += dp[i][1] - dp[i - 1][1];}printf("%lld %lld\n", ans1, ans2);}return 0;}

D

题意：给出数组a，n个数，q次操作，操作1 把区间l r的数向右平移一位，r位置上的数会到l位置上，操作2 反转区间l r，然后出一个数组b，求q次操作后的数组a中索引为b数组的所有数

思路：直接求数组a不好求，考虑到数组b的大小，我们可以将数组b给导回去，这样可以再m × q的复杂度内求出结果

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 2e5 + 10;const int INF = 1e9 + 10;int n, q, m;int a[qq], b[105];int p[qq], l[qq], r[qq];int main(){scanf("%d%d%d", &n, &q, &m);for(int i = 1; i <= n; ++i) {scanf("%d", a + i);}for(int i = 1; i <= q; ++i) {scanf("%d%d%d", p + i, l + i, r + i);}for(int i = 1; i <= m; ++i) {scanf("%d", b + i);}for(int i = q; i >= 1; --i) {for(int j = 1; j <= m; ++j) {if(l[i] <= b[j] && b[j] <= r[i]) {if(p[i] == 2) {b[j] = r[i] - b[j] + l[i];} else {if(b[j] == l[i])b[j] = r[i];elseb[j]--;}}}}for(int i = 1; i <= m; ++i) {printf("%d ", a[b[i]]);}puts("");return 0;}