Codeforces Round #443 (Div. 2)

A

比较傻用了二分

#include<bits/stdc++.h>using namespace std;#define LL long long#define mst(a, b)memset(a, b, sizeof a)#define pb push_backconst int qq = 1e7 + 10;const int MOD = 1e9 + 7;const int INF = 1e9 + 10;int vis[qq];int a[qq], b[qq];int main() {int n;scanf("%d", &n);mst(vis, 0);int start = -1;for(int i = 1; i <= n; ++i) {scanf("%d%d", a + i, b + i);if(start == -1) {start = a[i];} else {int l = 0, r = 1e6;int tmp;while(l <= r) {int mid = (l + r) >> 1;if(a[i] + mid * b[i] > start) {tmp = mid;r = mid - 1;} else {l = mid + 1;}}start = a[i] + tmp * b[i];}}printf("%d\n", start);return 0;}

B

很显然当k>=n的时候肯定就是数列中数最大的那个数的人赢，其他情况模拟下即可

#include<bits/stdc++.h>using namespace std;#define LL long long#define mst(a, b)memset(a, b, sizeof a)#define pb push_backconst int qq = 1e5 + 10;const int MOD = 1e9 + 7;const int INF = 1e9 + 10;int vis[505];int a[505];struct Node {int id, pw;}p;int main() {int n, k;scanf("%d%d", &n, &k);int maxn = -1, id = -1;for(int i = 1; i <= n; ++i) {scanf("%d", a + i);if(a[i] > maxn) {maxn = a[i];id = i;}}if(k > n) {printf("%d\n", maxn);return 0;}queue<Node> Q;if(a[2] > a[1]) {id = 2, maxn = a[2];} else {id = 1, maxn = a[1];}for(int i = 1; i <= n; ++i) {p.id = i, p.pw = a[i];if(id == i)continue;Q.push(p);}while(true) {p = Q.front();Q.pop();if(maxn > p.pw) {vis[id]++;Q.push(p);} else {Node tmp;tmp.pw = maxn, tmp.id = id;Q.push(tmp);maxn = p.pw;id = p.id;vis[id]++;}if(vis[id] == k) {printf("%d\n", maxn);return 0;}}}

C

我们知道如果进行 | 操作那么就可以确定某些位置肯定是1，如果进行 & 操作就可以确定某些位置肯定是0， 如果进行 ^ 操作如果是1的位置肯定是0，是0的位置肯定是1，未操作的位置可以覆盖 ^ 操作标记，如果本身已经被 ^ 操作覆盖那么这位重新变成未操作状态

#include<bits/stdc++.h>using namespace std;#define LL long long#define mst(a, b)memset(a, b, sizeof a)#define pb push_back#define mk make_pair#define fi first#define se second#define pill pair<int, int>const int qq = 1e5 + 10;const int MOD = 1e9 + 7;const int INF = 1e9 + 10;int bit[15];int main() {mst(bit, -1);int n;scanf("%d", &n);char op[6];int x;while(n--) {scanf("%s%d", op, &x);int cnt = 0;if(op[0] == '|') {while(x > 0) {if(x & 1)bit[cnt] = 1;cnt++;x >>= 1;}} else if(op[0] == '&') {x = 1023 - x;while(x > 0) {if(x & 1)bit[cnt] = 0;cnt++;x >>= 1;}} else {while(x > 0) {if(x & 1) {if(bit[cnt] == 1)bit[cnt] = 0;else if(bit[cnt] == 0)bit[cnt] = 1;else if(bit[cnt] == -1)bit[cnt] = 2;else if(bit[cnt] == 2)bit[cnt] = -1;}cnt++;x >>= 1;}}}int ans[3];ans[0] = 0, ans[1] = 1023, ans[2] = 0;for(int i = 0; i < 10; ++i) {if(bit[i] == 1)ans[0] += (1 << i);else if(bit[i] == 0)ans[1] -= (1 << i);else if(bit[i] == 2)ans[2] += (1 << i);}printf("3\n| %d\n& %d\n^ %d\n", ans[0], ans[1], ans[2]);return 0;}

D

题意：n个数字，讲这n个数字再重复m - 1次，删掉连续的k个相同的数，直到删除到不能再删为止，问最后剩下多少个数

思路：我们可以考虑相邻两个n个数字之间的删除情况，这两个确定了就可以确定之后的删除情况了，我们首先要预处理一下原本的n个数，也就是能删的就先删掉，具体看代码吧 能力有限

#include<bits/stdc++.h>using namespace std;#define LL long long#define mst(a, b)memset(a, b, sizeof a)#define pb push_back#define mk make_pair#define fi first#define se second#define pill pair<LL, LL>const int qq = 1e5 + 10;const int MOD = 1e9 + 7;const int INF = 1e9 + 10;LL n, k, m;pill p[qq];int main() {scanf("%lld%lld%lld", &n, &k, &m);int pre = -1;int top = 0, cnt = 0;p[0].fi = -1;for(int x, i = 0; i < n; ++i) {scanf("%d", &x);if(!top || p[top].fi != x) {p[++top].fi = x;p[top].se = 1;} else {p[top].se++;}if(p[top].se == k) {p[top--].se -= k;}}LL sum = 0;for(int i = 1; i <= top; ++i) {sum += p[i].se;}if(top == 0) {printf("0\n");return 0;}if(m == 1) {printf("%d\n", sum);} else if(k >= n) {if(top == 1) {printf("%lld\n", sum * m % k);} else {printf("%lld\n", sum * m);}} else if(k < n) {if(top == 1) {printf("%lld\n", sum * m % k);} else {int l = 1, r = top;LL d = 0, tmp = 0;while(l < r) {if(p[l].fi != p[r].fi)break;if((p[l].se + p[r].se) % k == 0) {d += p[l].se + p[r].se;l++, r--;} else {d += (p[l].se + p[r].se) - (p[l].se + p[r].se) % k;break;}}LL ans = 0;if(l < r) {printf("%lld\n", sum * m - d * (m - 1));} else if(l == r) {if(p[l].se * m % k == 0) {ans = 0;} else {ans = sum * m - d * (m - 1) - (p[l].se * m / k * k);}printf("%lld\n", ans);}}}return 0;}