Codeforces Round #320 (Div. 2) [Bayan Thanks-Round]
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;
}
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) { //UNAC
double 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;
}
编译人生，运行世界！
