#725-Div.3

A. Stone Game

先找到最大值最小值的位置，然后有三种选取（两边中走一边/两边都走一点）

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 105
int T, n;
int a[maxn];
bool vis[maxn];int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d", &n );for( int i = 1;i <= n;i ++ )scanf( "%d", &a[i] );int minn = maxn, maxx = -maxn, pos_min, pos_max;for( int i = 1;i <= n;i ++ ) {if( a[i] < minn ) minn = a[i], pos_min = i;		if( a[i] > maxx ) maxx = a[i], pos_max = i;}int l = min( pos_min, pos_max ), r = max( pos_min, pos_max );printf( "%d\n", min( r, min( n - l + 1, l + n - r + 1 ) ) );}return 0;
}

B. Friends and Candies

首先糖果要能被均分，然后肯定是糖果多的人分给别人

#include <cstdio>
#define maxn 200005
int T, n;
int a[maxn];int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d", &n );int tot = 0;for( int i = 1;i <= n;i ++ ) {scanf( "%d", &a[i] );tot += a[i];}if( tot % n ) printf( "-1\n" );else {tot /= n;int ans = 0;for( int i = 1;i <= n;i ++ )if( a[i] > tot ) ans ++;printf( "%d\n", ans );}}return 0;
}

C. Number of Pairs

先排个序，用lower_bound/upper_bound查询要求值域中的值

#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 200005
#define int long long
int T, n, l, r;
int a[maxn];signed main() {scanf( "%lld", &T );while( T -- ) {scanf( "%lld %lld %lld", &n, &l, &r );int ans = 0;for( int i = 1;i <= n;i ++ )scanf( "%lld", &a[i] );sort( a + 1, a + n + 1 );for( int i = 1;i < n;i ++ ) {if( a[i] > r ) continue;int L, R;R = upper_bound( a + i + 1, a + n + 1, r - a[i] ) - a - 1 - 1;L = lower_bound( a + i + 1, a + n + 1, l - a[i] ) - a - 1 - 1;ans += R - L;}printf( "%lld\n", ans );}return 0;
}

D. Another Problem About Dividing Numbers

特判k=1的情况

最小的操作次数肯定是$2$，直接除成$1$，所以只需要考虑最大操作次数，在这之间的操作次数肯定都是可以达到的，显然

最大操作次数就是唯一质因数分解的幂次和

刚开始还担心根号带$T$跑不过(#^.#)

#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;
int T, a, b, k;int gcd( int x, int y ) {if( x < y ) swap( x, y );if( ! y ) return x;else return gcd( y, x % y );
}int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d %d %d", &a, &b, &k );if( k == 1 ) {if( a == b ) printf( "No\n" );else if( a % b == 0 || b % a == 0 ) printf( "Yes\n" );else printf( "No\n" );continue;}int cnta = 0, Sqr = sqrt( a );for( int i = 2;i <= Sqr;i ++ )if( a % i == 0 )while( a % i == 0 ) a /= i, cnta ++;if( a != 1 ) cnta ++;int cntb = 0; Sqr = sqrt( b );for( int i = 2;i <= Sqr;i ++ )if( b % i == 0 )while( b % i == 0 ) b /= i, cntb ++;if( b != 1 ) cntb ++;if( k <= cnta + cntb ) printf( "Yes\n" );else printf( "No\n" );}return 0;
}

E. Funny Substrings

过分原始的暴力模拟不可取，疯狂自叠加字符串长度可以达到$2^{50}$级别，望而却步

事实上对于每一个字符串，我们只需要存下haha的个数，长度，以及前后各三个字符即可

因为只有这特殊的六个字符才可能与新的字符串相加时产生新的haha

#include <map>
#include <cstdio>
#include <iostream>
using namespace std;
#define int long long
struct word {int len, cnt;string s;
};
map < string, word > mp;string GetHead( string s ) {if( s.size() < 3 ) return s;else return s.substr( 0, 3 );
}string GetTail( string s ) {if( s.size() < 3 ) return s;else return s.substr( s.size() - 3, 3 );
}int calc( string s, string p ) {int cnt = 0;for( int i = 0;i + p.size() <= s.size();i ++ )if( s.substr( i, p.size() ) == p ) cnt ++;return cnt;
}word merge( word a, word b ) {word New;New.len = a.len + b.len;New.cnt = a.cnt + b.cnt + calc( GetTail( a.s ) + GetHead( b.s ), "haha" );New.s = a.s + b.s;if( New.s.size() > 7 ) New.s = GetHead( a.s ) + "#" + GetTail( b.s );return New;
}signed main() {int T, n;string a, b, c, op, plus;scanf( "%lld", &T );while( T -- ) {mp.clear();scanf( "%lld", &n );int ans = 0;while( n -- ) {cin >> a >> op >> b;if( op == ":=" ) {word New;New.cnt = calc( b, "haha" );New.len = b.length();New.s = b;mp[a] = New;}else {cin >> plus >> c;mp[a] = merge( mp[b], mp[c] );}ans = mp[a].cnt;}printf( "%lld\n", ans );}return 0;
}

F. Interesting Function

差分一下

发现对于任何数位上的值，完整向前一位进$1$，变换次数都是$10$

所以枚举数位，看前面进位次数乘个$10$，再加上该数位的值即可

#include <cstdio>
#define int long long
int cnt[20];int solve( int x ) {int ans = 0;while( x ) {int r = x % 10; x /= 10;ans += x * 10 + r;}return ans;
}signed main() {int T, l, r;scanf( "%lld", &T );while( T -- ) {scanf( "%lld %lld", &l, &r );printf( "%lld\n", solve( r ) - solve( l ) );}return 0;
}

G. Gift Set

能分成$n$个组合，就蕴含$n-1$

所以考虑二分组合个数$t$，假设$x>y,a>b$

$x,y$都至少会减去$t\times b$个，剩下了$t\times (a-b)$就随机让$x-t\times b,y-t\times b$补上

用向下取整判断即可

#include <cstdio>
#include <iostream>
using namespace std;
int T, x, y, a, b;bool check( int t ) {int X = x - t * b, Y = y - t * b;if( X / ( a - b ) + Y / ( a - b ) >= t ) return 1;else return 0;
}int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d %d %d %d", &x, &y, &a, &b );if( a == b ) {printf( "%d\n", min( x / a, y / b ) );continue;}if( x < y ) swap( x, y );if( a < b ) swap( a, b );int l = 0, r = y / b, ans;while( l <= r ) {int mid = ( l + r ) >> 1;if( check( mid ) ) ans = mid, l = mid + 1;else r = mid - 1;}printf( "%d\n", ans );}return 0;
}