NOIP模拟题 2017.11.6

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cctype>
#include <algorithm>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <set>
#include <map>
#ifdef WIN32
#define Auto "%I64d"
#else
#define Auto "%lld"
#endif
using namespace std;
typedef bool boolean;
typedef pair<int, int> pii;
#define ll long long
#define smin(_a, _b) _a = min(_a, _b)
#define smax(_a, _b) _a = max(_a, _b)

template<typename T>
inline void readInteger(T& u) {
char x;
while(!isdigit(x = getchar()));
for(u = x - '0'; isdigit(x = getchar()); u = u * 10 + x - '0');
}

int m, k;
int n = 0;
int counter[1005];
int X[1005], Y[1005], Z[1005], S;

int myrand() {
return rand() << 15 | rand();
}

inline void init() {
readInteger(m);
readInteger(k);
S = (1 << k) - 1;
for(int i = 1; i <= m; i++) {
readInteger(counter[i]);
n += counter[i];
}
for(int i = 1; i <= m; i++)
readInteger(X[i]);
for(int i = 1; i <= m; i++)
readInteger(Y[i]);
for(int i = 1; i <= m; i++)
readInteger(Z[i]);
}

int pos[10005];

inline void solve() {
int len = 0;
for(int i = 1; i <= m; i++) {
pos[++len] = X[i];
for(int j = 1; j <= counter[i] && j < 6; j++)
pos[++len] = (pos[len - 1] * Y[i] + Z[i]) & S;
}

pos[0] = -1;
int cmp = 0, id, maxcnt = 0, cnt, last;
for(int i = 1; i < len; i++) {
if(pos[i] != pos[i - 1])
cmp = 0;
if(++cmp > maxcnt)
maxcnt = cmp, id = pos[i];
}

cnt = 0;
for(int i = 1; i <= m; i++) {
cnt += (last = X[i]) == id;
for(int j = 1; j < counter[i]; j++)
cnt += (last = (last * Y[i] + Z[i]) & S) == id;
}

if(n - cnt >= cnt - 1)
printf("0");
else
printf("%d", 2 * cnt - n - 1);
}

int main() {
freopen("read.in", "r", stdin);
freopen("read.out", "w", stdout);
//    srand(233);
init();
solve();
return 0;
}

read (Random II)

Solution 3 (求和法)

　　因为它出现次数大于一半，所以考虑用一个 cnt 和一个 id

　　枚举序列中每个数，如果 cnt == 0 ，那么就将 id 赋值为当前枚举的这个数，并将cnt置为1。

　　否则，如果当前的这个数和 id 相等，就将 cnt 的值加1，否则减1。

　　这个算法完成后，我们会得到一个是出现次数超过n的一半的众数或者一个诡异的数，最后再把得到的id带回去求次数。

　　这么做的正确性显然(虽然解释不了但是觉得显然正确啊)。

Code

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cctype>
#include <algorithm>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <set>
#include <map>
#ifdef WIN32
#define Auto "%I64d"
#else
#define Auto "%lld"
#endif
using namespace std;
typedef bool boolean;
typedef pair<int, int> pii;
#define ll long long
#define smin(_a, _b) _a = min(_a, _b)
#define smax(_a, _b) _a = max(_a, _b)

template<typename T>
inline void readInteger(T& u) {
char x;
while(!isdigit(x = getchar()));
for(u = x - '0'; isdigit(x = getchar()); u = u * 10 + x - '0');
}

int m, k;
int n = 0;
int *counter;
int *X, *Y, *Z, S;

inline void init() {
readInteger(m);
readInteger(k);
counter = new int[(m + 1)];
X = new int[(m + 1)];
Y = new int[(m + 1)];
Z = new int[(m + 1)];
S = (1 << k) - 1;
for(int i = 1; i <= m; i++) {
readInteger(counter[i]);
n += counter[i];
}
for(int i = 1; i <= m; i++)
readInteger(X[i]);
for(int i = 1; i <= m; i++)
readInteger(Y[i]);
for(int i = 1; i <= m; i++)
readInteger(Z[i]);
}

int cnt = 0, id;

inline void add(int x) {
if(cnt == 0)
id = x, cnt = 1;
else if(id == x)
cnt++;
else
cnt--;
}

inline void solve() {
int last;
for(int i = 1; i <= m; i++) {
add(last = X[i]);
for(int j = 1; j < counter[i]; j++)
add(last = (last * 1LL * Y[i] + Z[i]) & S);
}

cnt = 0;
for(int i = 1; i <= m; i++) {
cnt += (last = X[i]) == id;
for(int j = 1; j < counter[i]; j++)
cnt += (last = (last * 1LL * Y[i] + Z[i]) & S) == id;
}

if(n - cnt >= cnt - 1)
printf("0");
else
printf("%d", 2 * cnt - n - 1);
}

int main() {
freopen("read.in", "r", stdin);
freopen("read.out", "w", stdout);
init();
solve();
return 0;
}

　　题目大意 (题目太简洁，无法概括大意)

　　因为涉及到了可恶的位运算，为了更好地处理它们，所以想到Trie树。

　　如果Trie树的一个非叶节点在两天中表示的名次在a ~ b之间，设它的两棵子树的大小分别为s1和s2。

　　那么左子树表示的区间就是a ~ (a + s1 - 1)和(a + s2) ~ b，右子树同理。

　　因为最终到了叶节点，表示的区间都变成a ~ a的形式，并且我们关心的只是平方和。

　　所以考虑如何维护所有开始端点的平方和。

　　写写式子发现:



　　由于然后发现再维护一下所有左端点的和就可以搞定了。

　　写代码的时候可以用黑科技优化，不建Trie树就可以直接搞答案。先将A数组排序，然后对于每一层都进行二分查找这一位0和1的分界位置。

Code

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cctype>
#include <algorithm>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <set>
#include <map>
#ifdef WIN32
#define Auto "%I64d"
#else
#define Auto "%lld"
#endif
using namespace std;
typedef bool boolean;
typedef pair<int, int> pii;
#define ll long long
#define smin(_a, _b) _a = min(_a, _b)
#define smax(_a, _b) _a = max(_a, _b)

template<typename T>
inline void readInteger(T& u) {
static char x;
while(!isdigit(x = getchar()));
for(u = x - '0'; isdigit(x = getchar()); u = u * 10 + x - '0');
}

const int M = 1e9 + 7;

int n, m;
int ans = 0;
ll S;
int* A;

inline void init() {
readInteger(n);
readInteger(m);
A = new int[(n + 1)];
S = (1ll << (m - 1));
for(int i = 1; i <= n; i++)
readInteger(A[i]);
}

void dfs(int dep, int L, int R, ll sum, ll sum2) {
if(L == R) {
ans ^= (sum2 % M);
return;
}
int l = L, r = R;
while(l <= r) {
int mid = (l + r) >> 1;
if(A[mid] & (1 << dep))    r = mid - 1;
else l = mid + 1;
}
int s1 = r - L + 1, s2 = R - r;
if(r >= L)    dfs(dep - 1, L, r, sum + S * s2, sum2 + (ll)sum * s2 + S * s2 * s2);
if(R > r)    dfs(dep - 1, r + 1, R, sum + S * s1, sum2 + (ll)sum * s1 + S * s1 * s1);
}

inline void solve() {
sort(A + 1, A + n + 1);
dfs(m - 1, 1, n, 0, 0);
printf("%d", ans);
}

int main() {
freopen("race.in", "r", stdin);
freopen("race.out", "w", stdout);
init();
solve();
return 0;
}