FZU 2105 Digits Count(AC)

因为0-15二进制下最大是1111，所以可以每个区间记录一下各个位上1的个数，这样不管是对于操作也好还是求和也好都简便了许多。

PS1.感谢薛神提点。

PS2.注意细节

#include <cstdio>
#include <cstdlib>
#include <cstring>

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

const int MAXN = 1000002;

int cnt[ MAXN << 2 ][4];
int flag[ MAXN << 2 ][4];
int Xor[ MAXN << 2 ][4];   //标记该位是否需要异或

int N, Q;

void PushUp( int rt )
{
    int lc = rt << 1;
    int rc = rt << 1 | 1;
    for ( int i = 0; i < 4; ++i )
        cnt[rt][i] = cnt[lc][i] + cnt[rc][i];
    return;
}

void PushDown( int rt, int m, int l, int r )
{
    int lc = rt << 1;
    int rc = rt << 1 | 1;
    for ( int i = 0; i < 4; ++i )
    {
        if ( flag[rt][i] != -1 )
        {
            flag[lc][i] = flag[rc][i] = flag[rt][i];
            cnt[lc][i] = flag[lc][i] * ( m - l + 1 );
            cnt[rc][i] = flag[rc][i] * ( r - m );
            flag[rt][i] = -1;
            Xor[lc][i] = Xor[rc][i] = 0;
        }
        else if ( Xor[rt][i] )
        {
            cnt[lc][i] = m - l + 1 - cnt[lc][i];
            cnt[rc][i] = r - m - cnt[rc][i];

            if ( flag[lc][i] != -1 ) flag[lc][i] ^= 1;
            else Xor[lc][i] ^= 1;

            if ( flag[rc][i] != -1 ) flag[rc][i] ^= 1;
            else Xor[rc][i] ^= 1;

            Xor[rt][i] = 0;
        }
    }
}

int Query( int L, int R, int l, int r, int rt )
{
    if ( L <= l && r <= R )
    {
        int sum = 0;
        for ( int i = 0; i < 4; ++i )
        {
            sum += ( 1 << i ) * cnt[rt][i];
       //     printf( "cnt[%d]=%d\n", i, cnt[rt][i] );
        }
        return sum;
    }

    int m = ( l + r ) >> 1;
    int ret = 0;

    PushDown( rt, m, l, r );

    if ( L <= m ) ret += Query( L, R, lson );
    if ( R > m )  ret += Query( L, R, rson );

    PushUp( rt );
    return ret;
}

void AND( int rt, int l, int r, int opn )
{
    for ( int i = 0; i < 4; ++i )
    {
        if ( ( 1 << i ) & opn ) continue;
        flag[rt][i] = 0;
        cnt[rt][i] = 0;
        Xor[rt][i] = 0;
    }
    return;
}

void OR( int rt, int l, int r, int opn )
{
    for ( int i = 0; i < 4; ++i )
    {
        if ( !( (1 << i) & opn ) ) continue;
        flag[rt][i] = 1;
        cnt[rt][i] = r - l + 1;
        Xor[rt][i] = 0;
    }
    return;
}

void XOR( int rt, int l, int r, int opn )
{
    for ( int i = 0; i < 4; ++i )
    {
        if ( !( ( 1 << i ) & opn ) ) continue;
        cnt[rt][i] = r - l + 1 - cnt[rt][i];
      //  printf( "**cnt[%d]=%d\n", i, cnt[rt][i] );
        if ( flag[rt][i] != -1 ) flag[rt][i] ^= 1;
        else Xor[rt][i] ^= 1;
    }
    return;
}

void Update( int L, int R, int c, int opn, int l, int r, int rt )
{
    if ( L <= l && r <= R )
    {
        if ( c == 1 ) AND( rt, l, r, opn );
        else if ( c == 2 ) OR( rt, l, r, opn );
        else XOR( rt, l, r, opn );

        return;
    }
    int m = ( l + r ) >> 1;
    PushDown( rt, m, l, r );

    if ( L <= m ) Update( L, R, c, opn, lson );
    if ( R > m ) Update( L, R, c, opn, rson );
    PushUp( rt );
    return;
}

void build( int l, int r, int rt )
{
    for ( int i = 0; i < 4; ++i )
    {
        flag[rt][i] = -1;
        Xor[rt][i] = 0;
    }

    if ( l == r )
    {
        int a;
        scanf( "%d", &a );
        for( int i = 0; i < 4; ++i )
            if ( a & ( 1 << i ) ) cnt[rt][i] = 1;
            else cnt[rt][i] = 0;
        return;
    }
    int m = ( l + r ) >> 1;
    build( lson );
    build( rson );
    PushUp( rt );
    return;
}

int main()
{
  //  freopen( "s.out", "w", stdout );
    int T;
    scanf( "%d", &T );
    while ( T-- )
    {
        scanf( "%d%d", &N, &Q );
        build( 0, N - 1, 1 );
        while ( Q-- )
        {
            int a, b, opn;
            char str[6];
            scanf( "%s", str );
            if ( str[0] == 'S' )
            {
                scanf( "%d%d", &a, &b );
                printf( "%d\n", Query( a, b, 0, N - 1, 1 ) );
            }
            else
            {
                scanf("%d%d%d", &opn, &a, &b );
                if ( str[0] == 'A' ) Update( a, b, 1, opn, 0, N - 1, 1 );
                else if ( str[0] == 'O' ) Update( a, b, 2, opn, 0, N - 1, 1 );
                else Update( a, b, 3, opn, 0, N - 1, 1 );
            }
        }
    }
    return 0;
}