Codeforces Round #262 (Div. 2)

A.Vasya and Socks

题意：有n个袜子，每天穿一双晚上扔掉，每m天再买一双。求一共能连续穿多少天？

题解：给n分成 n/m*m和n-n/m*m两部分递归计算。

B.Little Dima and Equation

题意：x = b·s(x)a + c，s(x)是x的各位数字和。输入a、b、c，求使等式成立的X。

题解：枚举每个S(x)，求对应的x。x的范围是0-81，不到为什么他们hack了好多72的。

C.Present

题意：有n盆花，每盆花的高度为a[i]。每天可以选一段连续的的序号的花浇灌，浇灌的花高度+1。连续浇m天，求最矮的花的高度。

题解：二分答案 + check()。用mi记录当前位置前面影响当前位置的浇灌次数，队列q记录影响的位置序列。当前位置需要高度大于mi，则当前位置需要浇灌mi-x天，mi+=x。若队首的与当前位置的距离大于w，则减去队首的浇灌次数。

代码：

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <deque>#include <set>#include <map>#include <string>using namespace std;#define For(i,a) for(i=0;i<a;i++)#define Foru(i,a,b) for(i=a;i<=b;i++)#define Ford(i,a,b) for(i=a;i>=b;i--)#define clr(ar,vel) memset(ar,vel,sizeof(ar))#define PB push_backtypedef long long ll;const int maxint = 0x7fffffff;const ll maxll = 1LL<<60;int a[100010];int b[100010];int n, m, w;queue<int> q;int check(ll x){//cout << x << endl;//cout << "###################" << endl;ll mi = 0;int cnt = 0, fn;while(q.size()) q.pop();for(int i = 0; i < n; i ++)b[i] = x-a[i];for(int i = 0; i < n; i ++){if( q.size()) {fn = q.front();if( i - fn >= w ) {mi -= b[fn];q.pop();}}//cout << mi << ' ' << b[i] << ' ' << i << ' ' << fn << endl;if( b[i] - mi > 0) {int t = b[i] - mi;mi += b[i]-mi;b[i] = t;q.push(i);cnt += b[i];}if( cnt > m ) return 0;}//cout << x << ' ' << cnt << endl; return cnt <= m;}int main(){ll sum;while(~scanf("%d%d%d",&n,&m,&w)){sum = 0;for(int i = 0; i < n; i ++) {scanf("%d",a+i);sum += a[i];}ll l, r, mid;l = 1; r = 2000000000;while( l < r){mid = (l+r)/2;if( check(mid)){//cout << "mid   " << mid << endl;l = mid+1;}else r = mid;}printf("%d\n",l-1);}return 0;}

D.Little Victor and Set

题意：输入l、r、k，求一个集合set，使得1 <= set.size <= k，队列元素连续异或的结果最小。

题解：1、k=1时，l最小。2、k=2时，i^(i+1)=1、3、5……，求最小值。3、k >= 4时，连续四个异或是0。

4、 k=3时，当时没想明白，看题解也看了半天才弄懂。i%2 == 0时，i^(i+1) == 1。若结果等于0，则三个数异或等于0。

1：3个数 l<=x<=y<=z<=r,因为x^y^z == 0,z的二进制最高位为1，则y的最高位为1，x的最高为等于0.2：若x的第二位为0，因为异或结果为0，则x,y的第二位同时为0或1.x、y、z更接近所以x,y的第二位为0。重复1.3：若x的第二位为1，因为异或结果为零，y<z，所以y的第二位为0，z的第二位为1.即：x: 011……11 -> mn <<= 1, mn++; y: 101……1 2 -> mx <<= 1, mx-1;z: 110……0 3 -> mx ;

代码：

<span style="font-size:12px;">#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <deque>#include <set>#include <map>#include <string>using namespace std;#define For(i,a) for(i=0;i<a;i++)#define Foru(i,a,b) for(i=a;i<=b;i++)#define Ford(i,a,b) for(i=a;i>=b;i--)#define clr(ar,vel) memset(ar,vel,sizeof(ar))#define PB push_backtypedef long long ll;const int maxint = 0x7fffffff;const ll maxll = 1LL<<60;int check3(ll l, ll r, ll k){ll mn = 1;ll mx = 3;while( mx <= r){if( mn >= l){cout << 0 << endl;cout << 3 << endl;cout << mn << ' ' << mx-1 << ' ' << mx << endl;return 0;}mn <<= 1;mx <<= 1;mn ++;}if( k >= 4) for(ll i = l+3; i <= r; i ++){if( (i^(i-1)^(i-2)^(i-3)) == 0){cout << 0 << endl;cout << 4 << endl;cout << i << ' ' << i-1 << ' ' << i-2 << ' ' << i-3 << endl;return 0;}}ll mi = l;for(ll i = l+1; i <= r; i ++){if( (i^(i-1)) == 1) {cout << 1 << endl;cout << 2 << endl;cout << i << ' ' << i-1 << endl;return 0;}if( (i^(i-1)) <= mi) mi = i^(i-1);}for(ll i = l+1; i <= r; i ++){if( (i^(i-1)) == mi) {cout << mi << endl;cout << 2 << endl;cout << i << ' ' << i-1 << endl;return 0;}}cout << l << endl;cout << 1 << endl;cout << l << endl;return 0;}int main(){ll n, m, k;int flag;while(cin >> n >> m >> k){flag = 0;if(k == 1 || m-n == 0) {cout << n << endl;cout << 1 << endl;cout << n << endl;}else if( k == 2 || m-n == 1){ll mi = n;for(ll i = n+1; i <= m; i ++){if( (i^(i-1)) == 1) {cout << 1 << endl;cout << 2 << endl;cout << i << ' ' << i-1 << endl;flag = 1;break;}if( (i^(i-1)) <= mi) mi = i^(i-1);}if( !flag ) for(ll i = n+1; i <= m; i ++){if( (i^(i-1)) == mi) {cout << mi << endl;cout << 2 << endl;cout << i << ' ' << i-1 << endl;flag = 1;break;}}if( !flag ) {cout << n << endl;cout << 1 << endl;cout << n << endl;}}else if( k >= 3 || m-n >= 2) {check3(n, m, k);}}return 0;}</span>

E.Roland and Rose

题意：

题解：