Codeforces Round #320 (Div. 2) [Bayan Thanks-Round]
2024-05-08 19:47:36
数学 A - Raising Bacteria
分析：如果1 << k == x，那么放1个就可以了；否则还要加上差值的二进制的1的个数。
/************************************************
* Author :Running_Time
* Created Time :2015/9/16 星期三 23:13:08
* File Name :A.cpp************************************************/#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;int main(void) {int x; scanf ("%d", &x);int i = 1;while (i < x) {i <<= 1;}if (i == x) {printf ("%d\n", 1);}else {int ans = 1;i >>= 1;int y = x - i;while (y) {if (y & 1) ans++;y >>= 1;}printf ("%d\n", ans);}return 0;
}
贪心 B - Finding Team Member
题意：每两个人组成一个team，两人之间有一个连接的强度，问所有人连接强度最大的team组合是怎样的
分析：按照连接强度排序，因为连接强度无重复。如果该连接强度连接的两个人没有组成team，那么他们就组成team，这样的选择是连接强度最大的
总结：这题的贪心方法没想到，按照每个人能连接的人的连接强度排序WA了
/************************************************
* Author :Running_Time
* Created Time :2015/9/16 星期三 23:13:11
* File Name :B.cpp************************************************/#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 8e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Team {int u, v, w;Team (int u = 0, int v = 0, int w = 0) : u (u), v (v), w (w) {}bool operator < (const Team &r) const {return w > r.w;}
}t[N*N];
int ans[N];int main(void) {int n; scanf ("%d", &n);int m = n * 2;int tot = 0;for (int i=2; i<=m; ++i) {for (int x, j=1; j<=i-1; ++j) {scanf ("%d", &x);t[++tot] = Team (i, j, x);}}sort (t+1, t+1+tot);memset (ans, 0, sizeof (ans));for (int i=1; i<=tot; ++i) {int u = t[i].u, v = t[i].v;if (!ans[u] && !ans[v]) {ans[u] = v; ans[v] = u;}}for (int i=1; i<=m; ++i) {printf ("%d%c", ans[i], i == m ? '\n' : ' ');}return 0;
}
二分 C - A Problem about Polyline
题意：有一个(0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – 的折线，问(a, b)是否在折线上
分析：两种情况，点在某个三角形的左边：a = 2 * k * x + b，或者在右边：a = 2 * k * x - b，满足x >= b，二分查找k
/************************************************
* Author :Running_Time
* Created Time :2015/9/16 星期三 23:13:14
* File Name :C.cpp************************************************/#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;int main(void) { //UNACdouble a, b;scanf ("%lf%lf", &a, &b);if (a < b) puts ("-1");else {double ans = 1e9 + 10;int l = 1, r = 1e9;while (l <= r) {int mid = (l + r) / 2;double xx = (a - b) / (double) (2 * mid);if (xx >= b) {ans = min (ans, xx);l = mid + 1;}else r = mid - 1;}l = 1, r = 1e9;while (l <= r) {int mid = (l + r) / 2;double xx = (a + b) / (double) (2 * mid);if (xx >= b) {ans = min (ans, xx);l = mid + 1;}else r = mid - 1;}printf ("%.9f\n", ans);}return 0;
}
贪心+位运算 D - "Or" Game
题意：可以对n个数字其中的数字*x，最多k次，问 
的最大值
分析：其决定因素的是二进制的最高位，如果ai*x，那么剩下的所有x都给ai，降低复杂度使用前缀和后缀，枚举选取ai得到最大值
/************************************************
* Author :Running_Time
* Created Time :2015/9/17 星期四 17:21:00
* File Name :D.cpp************************************************/#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
ll a[N], prefix[N], suffix[N];int main(void) {int n, k, x; scanf ("%d%d%d", &n, &k, &x);for (int i=1; i<=n; ++i) scanf ("%I64d", &a[i]);ll y = 1;for (int i=1; i<=k; ++i) y *= x;for (int i=1; i<=n; ++i) {prefix[i] = prefix[i-1] | a[i];}for (int i=n; i>=1; --i) {suffix[i] = suffix[i+1] | a[i];}ll ans = 0;for (int i=1; i<=n; ++i) {ans = max (ans, prefix[i-1] | (a[i] * y) | suffix[i+1]);}printf ("%I64d\n", ans);return 0;
}
三分||二分 E - Weakness and Poorness
题意：n个数字，每个数字减去x，使得求最大连续子序列和最小连续子序列的绝对值的最大值最小
分析：x增加，最大连续子序列f(x)递减，最小连续子序列g(x)递增，可用三分。二分的没想明白，先放着
三分：
/************************************************
* Author :Running_Time
* Created Time :2015/9/17 星期四 17:56:57
* File Name :E.cpp************************************************/#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e5 + 10;
const double INF = 1e10;
const int MOD = 1e9 + 7;
double a[N], b[N];
int n;double cal_max(void) {double sum = 0, mx = 0;for (int i=1; i<=n; ++i) {sum = max (sum + b[i], b[i]);if (sum > mx) mx = sum;}return mx;
}double cal_min(void) {double sum = 0, mn = INF;for (int i=1; i<=n; ++i) {sum = min (sum + b[i], b[i]);if (sum < mn) mn = sum;}return mn;
}double cal(double x) {for (int i=1; i<=n; ++i) b[i] = a[i] - x;return max (cal_max (), -cal_min ());
}int main(void) {scanf ("%d", &n);for (int i=1; i<=n; ++i) scanf ("%lf", &a[i]);double l = -INF, r = INF;for (int i=1; i<=200; ++i) {double mid = (l + r) / 2;double lmid = (l + mid) / 2;double rmid = (r + mid) / 2;if (cal (lmid) < cal (rmid)) r = rmid;else l = lmid;}printf ("%.10f\n", cal ((l + r) / 2));return 0;
}
二分：
/************************************************
* Author :Running_Time
* Created Time :2015/9/17 星期四 17:56:57
* File Name :E.cpp************************************************/#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e5 + 10;
const double INF = 1e10;
const int MOD = 1e9 + 7;
double a[N], b[N];
int n;double cal_max(void) {double sum = 0, mx = 0;for (int i=1; i<=n; ++i) {sum = max (sum + b[i], b[i]);if (sum > mx) mx = sum;}return mx;
}double cal_min(void) {double sum = 0, mn = INF;for (int i=1; i<=n; ++i) {sum = min (sum + b[i], b[i]);if (sum < mn) mn = sum;}return mn;
}bool check(double x) {for (int i=1; i<=n; ++i) b[i] = a[i] - x;double A = cal_max ();double B = cal_min ();return (-B > A);
}int main(void) {scanf ("%d", &n);for (int i=1; i<=n; ++i) scanf ("%lf", &a[i]);double l = -INF, r = INF;for (int i=1; i<=200; ++i) {double mid = (l + r) / 2;if (check (mid)) r = mid;else l = mid;}for (int i=1; i<=n; ++i) b[i] = a[i] - l;printf ("%.10f\n", cal_max ());return 0;
}转载于:https://www.cnblogs.com/Running-Time/p/4818822.html
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