P1226 [template] rapid power modulo operation ||
Quick question the power of the board:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef unsigned long long ll; 4 ll Mul ( ll a ,ll b , ll p ){ 5 a %= p; 6 b %= p; 7 ll c = (long double ) a*b / p; 8 ll ans = a*b - c*p; 9 if( ans < 0 ) ans += p ; 10 else if( ans >= p ) ans -= p ; 11 return ans ; 12 } 13 ll qmul( ll a , ll b , ll p ){ 14 ll ans = 0; 15 a %= p ; 16 while ( b ){ 17 if( b & 1 ){ 18 ans = (ans + a)%p; 19 } 20 a =( a + a )%p; 21 b>>=1; 22 } 23 return ans ; 24 } 25 ll qpow ( ll a , ll b , ll mod ){ 26 a %= mod ; 27 ll ans = 1 ; 28 while ( b ){ 29 if( b & 1 ){ 30 ans = qmul( ans , a , mod ); 31 } 32 a = qmul(a,a,mod); 33 b >>= 1 ; 34 } 35 return ans % mod ; 36 } 37 int main() 38 { 39 ll a,b,p,ans; 40 cin >> a >> b >> p ; 41 ans = qpow(a,b,p); 42 cout<<a<<"^"<<b<<" mod "<<p<<"="<<ans<<endl; 43 return 0 ; 44 }
A power of P1010
【answer】
Used is based dfs to do the practice, unless the number is on the current position of the digit is 1 or 2, or direct the next level, should set out a "2 (" Before entering the next level, out of time, cover go back" ) ".
1 #include<bits/stdc++.h> 2 using namespace std; 3 void dfs( int x ){ 4 if( x == 0 ){ 5 printf("2(0)"); 6 }else if( x == 1 ){ 7 printf("2"); 8 }else{ 9 bool f = false ; 10 for( int i=14 ; i>=0 ; i-- ){ 11 if( ( x & (1 << i) ) == (1 << i) ){ 12 if( f ) printf("+"); 13 else f = true; 14 15 if( i != 1 && i != 0 ){ 16 printf("2("); 17 dfs( i ) ; 18 printf(")"); 19 }else{ 20 dfs(i); 21 } 22 } 23 } 24 } 25 } 26 int main() 27 { 28 int n; 29 scanf("%d",&n); 30 dfs(n); 31 puts(""); 32 return 0; 33 }
P1908 reverse order
【answer】
[Fenwick tree]
We use two methods to try first is the Fenwick tree, this is not to say, the basic routine operations, that is, after discretization, maintaining each position has approached a number of buckets, now with the current index - calculated on the number of reverse positions corresponding form in front of the number of buckets, the rest as well.
[Merge sort]
Use merge sort to do, this problem is even more appropriate.
First minutes after treatment.
Mainly in terms of governance, two sorted arrays, pointers pointing to the location of the two heads is set to p1, p2.
p1 <p2: directly put to
p1> p2: this time requires array foregoing statistics, the number of previous remaining array.
Fenwick tree approach:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N = 5e5+100; 5 6 ll c[N]; 7 8 typedef struct Node { 9 int id , val ; 10 }Node ; 11 12 bool cmp1 ( Node u , Node v ){ 13 if( u.val == v.val ) return u.id < v.id ; 14 return u.val < v.val ; 15 } 16 bool cmp2 ( Node u , Node v ){ 17 return u.id < v.id ; 18 } 19 20 int lowbit( int x ){ 21 return x & -x ; 22 } 23 24 void update( int No , int val ){ 25 while( No < N ){ 26 c[No] += val ; 27 No += lowbit(No); 28 } 29 } 30 31 ll getsum( int No ){ 32 ll res = 0; 33 while( No > 0 ){ 34 res += c[No]; 35 No -= lowbit(No); 36 } 37 return res ; 38 } 39 Node a[N]; 40 int main() 41 { 42 int n; 43 scanf("%d",&n); 44 for( int i = 1 ; i <= n; i++){ 45 scanf("%d",&a[i].val); 46 a[i].id = i ; 47 } 48 sort( a+1 , a+1+n , cmp1 ); 49 for( int i=1;i<=n;i++){ 50 a[i].val = i ; 51 } 52 sort( a+1 , a+1+n , cmp2 ); 53 ll ans = 0 ; 54 for( int i=1 ; i <= n ;i++ ){ 55 ans += i - getsum(a[i].val)-1 ; 56 update( a[i].val , 1 ); 57 } 58 printf("%lld\n",ans); 59 return 0 ; 60 }
Merge sort practice:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N = 5e5+100; 5 ll a[N],b[N]; 6 ll ans = 0 ; 7 void Merge_sort( int L , int R ){ 8 if( L == R ) return ; 9 int Mid = L + R >> 1 ; 10 11 Merge_sort( L , Mid ) ; 12 Merge_sort( Mid+1 , R ); 13 14 int i = L , j = Mid + 1 , k = L ; 15 while( i <= Mid && j <= R ){ 16 if( a[i] <= a[j] ){ 17 b[k++] = a[i++]; 18 }else{ 19 b[k++] = a[j++]; 20 ans += Mid - i + 1 ; 21 } 22 } 23 while( i <= Mid ) b[k++] = a[i++]; 24 while( j <= R ) b[k++] = a[j++]; 25 for( int i = L ; i <= R ; i++ ){ 26 a[i] = b[i] ; 27 } 28 } 29 int main() 30 { 31 int n; 32 scanf("%d",&n); 33 for( int i=1 ;i<=n;i++){ 34 scanf("%lld",&a[i]); 35 } 36 Merge_sort( 1 , n ); 37 printf("%lld\n",ans); 38 return 0; 39 }
P1498 Nanban totem
【answer】
When you find the law, this problem is solved
The first is to create a process first, then pan to the right, to the right under the pan, then finished. The main backwards to come.
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N = 2e3+500; 5 int a[N>>1][N]; 6 int main() 7 { 8 int n; 9 scanf("%d",&n); 10 11 for( int i = 0 ; i < N>>1 ; i++ ){ 12 for( int j = 0 ; j < N ; j++ ){ 13 a[i][j] = ' '; 14 } 15 } 16 17 int len = 4 ; 18 a[0][0] = a[1][1] = '/'; 19 a[0][1] = a[0][2] = '_'; 20 a[0][3] = a[1][2] = '\\'; 21 22 while( --n ){ 23 for( int i = len/2 ; ~i ; i-- ){ 24 for( int j = 0 ; j < len ; j ++ ){ 25 a[i][j+len] = a[i+len/2][j+len/2] = a[i][j]; 26 } 27 } 28 len <<= 1 ; 29 } 30 31 for( int i = len/2 -1 ; ~i ; i-- ){ 32 for( int j = 0 ; j < len + (len-i) ; j ++ ){ 33 printf("%c",a[i][j]); 34 } 35 puts(""); 36 } 37 return 0 ; 38 }