Tutorials for 2014 SWJTU ACM Summer Training Team-PK Contest #1
A题：分段筛法+二分搜索。首先预处理出区间[1, 10^8]中的所有素数，然后对于查询区间[a, b]，只需要用区间[1, b]中的素数个数减去区间[1, a - 1]中的素数个数，显然可以二分得到。在筛选素数的过程中，并不是直接筛，而是采用分段筛法，即将区间[1, 10^8]划分为50个不相交的等长区间，分别筛选即可。
参考代码：
//#define SPJ
#ifdef SPJ
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#endif
#define dbg(x) cout << #x << " = " << x << endl
#define dbg2(x,y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
#define dbg3(x,y,z) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl
#define out(x) cout << (x) << endl
#define out2(x,y) cout << (x) << " " << (y) << endl
int prime[6000000]; int tot_prime = 0;
bool vst[2000000 + 10];
void Get_prime_1(int M)
{/*
memset(vst, 0, sizeof(vst));
int i, j, k, Sq = 1 + (int)sqrt((M << 1) + 1.0);
for(i = 3; i <= Sq; i += 2) {
if( vst[i >> 1] ) continue;
for(j = i * i, k = (i << 1); j <= ((M << 1) | 1); j += k)
vst[j >> 1] = 1;
}
for(i = 1; i <= M; i ++)
if( !vst[i] )
prime[tot_prime ++] = (i << 1) | 1;*/
int i, j;
memset(vst, 0, sizeof(vst));
for(i = 3; i <= M; i += 2) {
if( vst[i] ) continue;
prime[tot_prime ++] = i;
for(j = i << 1; j <= M; j += i)
vst[j] = 1;
}
}
void Get_prime_2(int l, int r)
{/*
memset(vst, 0, sizeof(vst));
int i, j, k;
for(i = 1; prime[i] * prime[i] <= ((r << 1) | 1); i ++) {
j = ((l << 1) | 1) / prime[i]; if( j % 2 == 0 ) j ++;
j = j * prime[i];
for(k = prime[i] << 1; j <= ((r << 1) | 1); j += k)
vst[(j >> 1) - l] = 1;
}
for(i = l; i <= r; i ++) {
if( !vst[i - l] && tot_prime < 5900000 )
prime[tot_prime ++] = ((i << 1) | 1);
}*/
memset(vst, 0, sizeof(vst));
int i, j;
for(i = 0; prime[i] * prime[i] <= r; i ++) {
j = l / prime[i] * prime[i];
if( j < l ) j += prime[i];
for(; j <= r; j += prime[i]) vst[j - l] = 1;
}
for(i = l; i <= r; i ++) {
if( !vst[i - l] ) {
prime[tot_prime ++] = i;
}
}
}
void init()
{
tot_prime = 0; prime[tot_prime ++] = 2;
Get_prime_1(2000000);
//printf("1: tot_prime = %d\n", tot_prime);
for(int i = 2; i <= 50; i ++) {
Get_prime_2(2000000 * (i - 1) + 1, 2000000 * i);
//printf("tot_prime = %d\n", tot_prime);
}
//dbg(tot_prime);
}
int main()
{
//freopen("data.in", "r", stdin);
//freopen("data.out", "w", stdout);
init(); int N, a, b, l, r, m, pos, pos_2;
while( scanf("%d %d", &a, &b) == 2 ) {
N = b;
l = 0, r = tot_prime - 1, pos = -1;
while( l <= r ) {
m = (l + r) >> 1;
if( prime[m] <= N ) pos = m, l = m + 1;
else r = m - 1;
}
N = a - 1;
l = 0, r = tot_prime - 1, pos_2 = -1;
while( l <= r ) {
m = (l + r) >> 1;
if( prime[m] <= N ) pos_2 = m, l = m + 1;
else r = m - 1;
}
printf("%d\n", (pos + 1) - (pos_2 + 1));
}
return 0;
}
B题：线段树+动态规划。容易推导出动态规划递推表达式，即DP[i] = max{DP[j] + A[i]}, (A[j] < A[i], 1 <= j < i)。但是这样做复杂度相当高。此时，我们需要优化，我们可以先对数组的值进行非递减排序，相同的值按照索引的大小排序（这一点很有必要）。然后，建立以索引为区间的线段树，进行动态更新和查询。
参考代码：
//#define SPJ
#ifdef SPJ
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#endif
#define dbg(x) cout << #x << " = " << x << endl
#define dbg2(x,y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
#define dbg3(x,y,z) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl
#define out(x) cout << (x) << endl
#define out2(x,y) cout << (x) << " " << (y) << endl
const int maxN = 100000 + 2;
typedef long long i64d;
struct Seq
{
int idx; i64d val;
inline void in(int &_idx) { scanf("%I64d", &val); idx = _idx; }
inline bool operator<(const Seq &s) const {
if( val != s.val ) return val < s.val;
return idx > s.idx;
}
};
Seq A[maxN]; int nA;
struct node
{
int l, r; i64d val;
};
node tree[maxN * 5];
void build(int l, int r, int idx)
{
tree[idx].l = l; tree[idx].r = r;
tree[idx].val = 0LL;
if( l == r ) return ;
int Mid = (l + r) >> 0x1;
build(l, Mid, idx << 0x1); build(1 + Mid, r, (idx << 0x1) + 1);
}
void update(int idx, int &ip, i64d &val)
{
if( tree[idx].l == tree[idx].r ) {
tree[idx].val = val; return ;
}
int Mid = (tree[idx].l + tree[idx].r) >> 0x1;
if( ip <= Mid ) update(idx << 0x1, ip, val);
else update(1 + (idx << 0x1), ip, val);
tree[idx].val = max(tree[idx << 0x1].val, tree[(idx << 0x1) + 1].val);
}
i64d query(int l, int r, int idx)
{
if( l == tree[idx].l && tree[idx].r == r )
return tree[idx].val;
int Mid = (tree[idx].l + tree[idx].r) >> 0x1;
if( r <= Mid ) return query(l, r, idx << 0x1);
else if( Mid < l ) return query(l, r, (idx << 0x1) + 1);
else return max(query(l, Mid, idx << 0x1), query(1 + Mid, r, (idx << 0x1) + 1));
}
i64d dp[maxN];
int main()
{
//freopen("data.in", "r", stdin);
//freopen("data.out", "w", stdout);
while( scanf("%d", &nA) == 1 ) {
for(int i = 0; i < nA; ++i) A[i].in(i);
sort(A, A + nA);
build(0, nA - 1, 1);
i64d tmp;
for(int i = 0; i < nA; ++i) {
tmp = query(0, A[i].idx, 1);
//printf("tmp = %lld, idx = %d\n", tmp, A[i].idx);
dp[i] = A[i].val + tmp;
update(1, A[i].idx, dp[i]);
}
i64d ans = dp[0];
for(int i = 1; i < nA; ++i) if( dp[i] > ans ) ans = dp[i];
printf("%I64d\n", ans);
}
return 0;
}
C题：推导题。这题有同学预先对字符串排序，失策没有卡死。实际上只需要统计每个字符的频率，然后从a到z获取频率为奇数的字符，并把得到的字符串中的最后一个字符去掉，就是答案（如果是空串就输出0）。
//#define SPJ
#ifdef SPJ
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#endif
#define dbg(x) cout << #x << " = " << x << endl
#define dbg2(x,y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
#define dbg3(x,y,z) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl
#define out(x) cout << (x) << endl
#define out2(x,y) cout << (x) << " " << (y) << endl
const int maxN = 1000000 + 10;
char str[maxN];
int cnt[26];
int main()
{
//freopen("data.in", "r", stdin);
//freopen("data.out", "w", stdout);
int i, j;
while( scanf("%s", str) == 1 ) {
memset(cnt, 0, sizeof(cnt));
for(i = 0; str[i]; i ++)
cnt[ str[i] - 'a' ] ++;
for(i = 25; i >= 0; i --)
if( cnt[i] & 1 ) {
cnt[i] ++;
break;
}
j = 0;
for(i = 0; i < 26; i ++)
if( cnt[i] & 1 ) {
j = 1;
putchar('a' + i);
}
if( j == 0 ) putchar('0');
putchar('\n');
}
return 0;
}
D题：推导题。有内存限制，意味着我们不能用数组存储，如果发现输入数据是64比特，n的规模又大，意味着我们往往需要使用getchar读入数据。接下来有两种思想，我实现了第2个，不知道有没有人用第1个，或者第1个会被卡掉？
第1种思想：容易在读入数据的过程中，得到3个数，即min（最小的数），max（最大的数），sum（和），显然答案等于 sum - (max - min + 1) * (max + min) / 2，小数组高精度处理即可。
第2种思想：同样的，我们可以在输入的过程中得到3个数，即min（最小的数），max（最大的数），XOR（所有数的异或值），显然答案等于 XOR ^ [min ^ (min + 1) ^ ... ^ max]，这里 ^ 表示异或运算。此时问题变成了需要求区间[min, max]所有的数的异或值，这是OJ上的一道陈题（链接），这里简单说下一种求解区间[a, b]所有的数的异或值的解法：
假设x是偶数，那么x ^ (x+1) 就等于1，也即x和(x+1)只有最低位不同。好了，我们会发现区间[a, b]是由大量的这样的x和x+1构成（除了边界情况）。
所以我们只需要从a和b的奇偶性入手即可，这里只考虑其中一种情况，即a是奇数，b是偶数，那么我们总可以把区间[a, b]划分为：
a, (a+1, a+2), (a+3, a+4), ..., (b-2, b - 1), b
显然中间有括号的两个数的异或值为1，假设这样的配对的数有y个，那么答案就是a ^ (y & 1) ^ b，不是吗？（其他情况可以参考着来。）
参考代码：
//#define SPJ
#ifdef SPJ
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#endif
#define dbg(x) cout << #x << " = " << x << endl
#define dbg2(x,y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
#define dbg3(x,y,z) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl
#define out(x) cout << (x) << endl
#define out2(x,y) cout << (x) << " " << (y) << endl
unsigned long long gao(unsigned long long l, unsigned long long r)
{
if( l == r )
return l;
unsigned long long res = 0;
if( l & 1LL ) {
res ^= l;
l ++;
}
if( !(r & 1LL) ) {
res ^= r;
r --;
}
if( l > r )
return res;
unsigned long long last = ((r - l + 1LL) >> 1) & 1LL;
return (res ^ last);
}
int main()
{
//freopen("data.in", "r", stdin);
//freopen("data.out", "w", stdout);
int n, i;
unsigned long long mi, ma, res, ans;
unsigned long long val;
char ch;
while( scanf("%d", &n) == 1 ) {
//scanf("%llu", &val);
while( (ch = getchar()) == ' ' || ch == '\n' ) ;
val = ch - '0';
while( (ch = getchar()) >= '0' && ch <= '9' )
val = val * 10 + (ch - '0');
ans = mi = ma = val;
for(i = 1; i < n; i ++) {
//scanf("%llu", &val);
while( (ch = getchar()) == ' ' || ch == '\n' ) ;
val = ch - '0';
while( (ch = getchar()) >= '0' && ch <= '9' )
val = val * 10 + (ch - '0');
if( mi > val )
mi = val;
if( ma < val )
ma = val;
ans ^= val;
}
res = gao(mi, ma);
//dbg3(mi, ma, res);
printf("%llu\n", (res ^ ans));
}
return 0;
}
E题：卡平方题。首先数据量大，请用getchar；其次再去计算每个数出现的次数，此时你可能会用平方（即 k * 100 *100）的复杂度去统计数据，如果输入数据有10^5组小数据，就很危险，可以先预处理，对付10^5组小数据的复杂度消耗是10^8数量级。（下一次可能有一道变型的题，请关注此题。）
参考代码：
//#define SPJ
#ifdef SPJ
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#endif
#define dbg(x) cout << #x << " = " << x << endl
#define dbg2(x,y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
#define dbg3(x,y,z) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl
#define out(x) cout << (x) << endl
#define out2(x,y) cout << (x) << " " << (y) << endl
const int maxN = 100 + 2;
vector<int> adj[maxN];
int sz[maxN];
int gcd(int a, int b)
{
if( b == 0 )
return a;
return gcd(b, a % b);
}
void init()
{
for(int i = 100; i >= 1; i --) {
for(int j = i; j >= 1; j --)
if( gcd(i, j) == 1 ) {
adj[i].push_back(j);
}
sz[i] = (int)adj[i].size();
}
}
int n, val, i, j, k;
long long times[maxN];
long long res, tmp;
char ch;
int main()
{
//freopen("data.in", "r", stdin);
//freopen("data.out", "w", stdout);
init();
while( scanf("%d", &n) == 1 ) {
memset(times, 0, sizeof(times));
k = 0;
for(i = 0; i < n; i ++) {
//scanf("%d", &val);
while( (ch = getchar()) == ' ' || ch == '\n' ) ;
val = ch - '0';
while( (ch = getchar()) >= '0' && ch <= '9' )
val = val * 10 + (ch - '0');
times[val] ++;
if( k < val ) k = val;
}
res = times[1] * times[1];
for(i = k; i > 1; i --) {
if( !times[i] ) continue;
tmp = 0LL; times[i] <<= 1;
for(j = 0; j < sz[i]; j ++)
tmp += times[ adj[i][j] ];
res += times[i] * tmp;
}
printf("%lld\n", res);
}
return 0;
}
（备注：所有比赛中都是解出0题最后的实习成绩并不一定就是不及格。还有其他成绩，比如100道题要求、出勤分等等，完成了集训的定量基本要求，肯定及格，不必太担心。）