FZU Problem 2105 Digits Count
Problem Description
Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations:
Operation 1: AND opn L R
Here opn, L and R are integers.
For L≤i≤R, we do A[i]=A[i] AND opn (here "AND" is bitwise operation).
Operation 2: OR opn L R
Here opn, L and R are integers.
For L≤i≤R, we do A[i]=A[i] OR opn (here "OR" is bitwise operation).
Operation 3: XOR opn L R
Here opn, L and R are integers.
For L≤i≤R, we do A[i]=A[i] XOR opn (here "XOR" is bitwise operation).
Operation 4: SUM L R
We want to know the result of A[L]+A[L+1]+...+A[R].
Now can you solve this easy problem?
Input
The first line of the input contains an integer T, indicating the number of test cases. (T≤100)
Then T cases, for any case, the first line has two integers n and m (1≤n≤1,000,000, 1≤m≤100,000), indicating the number of elements in A and the number of operations.
Then one line follows n integers A[0], A[1], ..., A[n-1] (0≤A[i]<16,0≤i<n).
Then m lines, each line must be one of the 4 operations above. (0≤opn≤15)
Output
Sample Input
Sample Output
Hint
A = [1 2 4 7]
SUM 0 2, result=1+2+4=7;
XOR 5 0 0, A=[4 2 4 7];
OR 6 0 3, A=[6 6 6 7];
SUM 0 2, result=6+6+6=18.
代码如下:
#include <stdio.h>
#include <string.h>
const int N = ; int c, res, a[N*][], flag[N*][], len[N*]; void pushDown( int rt )
{
int ls = * rt; int rs = ls | ;
for( int i = ; i < ; ++i )
{
if( flag[rt][i] == - ) continue;
if( flag[rt][i] == || flag[rt][i] == )
{
a[ls][i] = ( flag[ls][i] = flag[rt][i] ) * len[ls];
a[rs][i] = ( flag[rs][i] = flag[rt][i] ) * len[rs];
flag[rt][i] = -;
}
else if( flag[rt][i] == )
{
flag[rt][i] = -;
}
else
{
a[ls][i] = len[ls] - a[ls][i];
if( flag[ls][i] == - ) flag[ls][i] = ;
else flag[ls][i] ^= ;
a[rs][i] = len[rs] - a[rs][i];
if( flag[rs][i] == - ) flag[rs][i] = ;
else flag[rs][i] ^= ;
flag[rt][i] = -;
} }
} void pushUp( int rt )
{
int ls = * rt; int rs = ls | ;
for( int i = ; i < ; ++i )
a[rt][i] = a[ls][i] + a[rs][i];
} void build( int l, int r, int rt )
{
len[rt] = r - l + ;
if( l == r )
{
scanf("%d", &c);
for( int i = ; i < ; ++i )
a[rt][i] = ( & (c>>i) );
}
else
{
int mid = ( l + r ) / ;
build(l, mid, * rt);
build(mid + , r, * rt + );
pushUp( rt );
}
} void OR( int rt, int opn )
{
for( int i = ; i < ; ++i )
{
if( !( (opn>>i) & ) ) continue;
flag[rt][i] = ;
a[rt][i] = len[rt];
}
} void ADD( int rt, int opn )
{
for( int i = ; i < ; ++i )
{
if( (opn>>i) & ) continue;
flag[rt][i] = ;
a[rt][i] = ;
}
} void XOR( int rt, int opn )
{
for( int i = ; i < ; ++i )
{
if( !( (opn>>i) & ) ) continue;
if( flag[rt][i] == - ) flag[rt][i] = ;
else flag[rt][i] ^= ;
a[rt][i] = len[rt] - a[rt][i];
}
} void query( int l, int r, int rt, const int aa, const int bb )
{
if( aa <= l && r <= bb )
{
res = res + a[rt][] + a[rt][] * + a[rt][] * + a[rt][] * ;
return;
}
pushDown( rt );
int mid = ( l + r ) / ;
int ls = * rt; int rs = ls | ;
if( mid >= aa ) query( l, mid, ls, aa, bb );
if( mid < bb ) query( mid + , r, rs, aa, bb );
} void update( int l, int r, int rt, const int aa, const int bb, const int opn, const int f )
{
if( aa <= l && r <= bb )
{
switch( f )
{
case : OR( rt, opn ); break;
case : ADD( rt, opn ); break;
case : XOR( rt, opn ); break;
}
return;
}
pushDown( rt );
int mid = ( l + r ) / ;
int ls = * rt; int rs = ls | ;
if( mid >= aa ) update( l, mid, ls, aa, bb, opn, f );
if( mid < bb ) update( mid + , r, rs, aa, bb, opn, f );
pushUp( rt );
} int main()
{
int t, n, m, aa, bb, opn; char cmd[];
scanf("%d", &t);
while( t-- )
{
memset( flag, -, sizeof(flag) );
scanf("%d%d", &n, &m);
build(, n-, );
while( m-- )
{
scanf("%s", cmd);
if( cmd[] == 'S' )
{
scanf("%d%d", &aa, &bb);
res = ;
query( , n - , , aa, bb );
printf("%d\n", res);
}
else
{
scanf("%d%d%d", &opn, &aa, &bb);
switch( cmd[] )
{
case 'O': update(, n-, , aa, bb, opn, ); break;
case 'A': update(, n-, , aa, bb, opn, ); break;
case 'X': update(, n-, , aa, bb, opn, ); break;
default : break;
}
}
}
}
return ;
}
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