树状数组题目

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树状数组

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POJ2352
树状数组求顺序对

# include 
# include 
# include 
# include 
# define LL long long
using namespace std ;
int lowbit (int x) {
    return x&(-x);
}
struct h{
	int l;
	int r;
};
int num[15010];
h a[15010];
int n;
int sum[42000];
void add(int p){
    while(p<=32001){
    	sum[p]++;
    	p+=lowbit(p);
    }
}
int ask(int p){
	int ans=0;
	while(p>0){
		ans+=sum[p];
		p-=lowbit(p);
	}
	return ans;
}

int main(){
	while(~scanf("%d",&n)){
		for(int i=1;i<=n;i++){
			scanf("%d%d",&a[i].l,&a[i].r);
			a[i].l++;
		}
		for(int i=1;i<=n;i++){	
			int ans=ask(a[i].l);
			add(a[i].l);
			num[ans]++;
		}
		for(int i=0;i

OpenJ_Bailian 2299
树状数组求逆序对

# include 
# include 
# include 
# include 
# define LL long long
using namespace std ;
const int N = 5e5 + 10 ;
int a [N] , b [N] , c [N] , n ;
int lowbit ( int x ) {
    return ( - x ) & x ;
}
void modify ( int pos , int val ) {
    for ( int i = pos ; i <= n ; i += lowbit ( i ) ) 
        c [i] += val ;
}
int query ( int pos ) {
    int ret = 0 ;
    for ( int i = pos ; i >= 1 ; i -= lowbit ( i ) ) 
        ret += c [i] ;
    return ret ;
}
int main () {
    while ( scanf ( "%d" , & n ) != EOF && n ) {
        memset ( c , 0 , sizeof c ) ;
        for ( int i = 1 ; i <= n ; ++ i ) 
		scanf ( "%d" , & a [i] ) , b [i] = a [i] ;
        sort ( a + 1 , a + 1 + n ) ;
        for ( int i = 1 ; i <= n ; ++ i ) 
		b [i] = lower_bound ( a + 1 , a + 1 + n , b [i] ) - a ;
        LL ans = 0 ;
        for ( int i = 1 ; i <= n ; ++ i ) {
            modify ( b [i] , 1 ) ; 
            ans += i - query ( b [i] ) ;
        }
        printf ( "%lld\n" , ans ) ; 
    }
    return 0 ;
}