其实就是找一个最小的level k使得level 1到k的ant数量和不小于rank。
树状数组：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 100001; 7 int a[N]; 8 int c[N]; 9 int n, m; 10 11 int lb( int i ) 12 { 13 return i & -i; 14 } 15 16 void update( int i, int v ) 17 { 18 while ( i <= n ) 19 { 20 c[i] += v; 21 i += lb(i); 22 } 23 } 24 25 int kth( int k ) 26 { 27 int ans = 0, cnt = 0; 28 for ( int i = 16; i >= 0; i-- ) 29 { 30 ans += ( 1 << i ); 31 if ( ans >= n || cnt + c[ans] >= k ) 32 { 33 ans -= ( 1 << i ); 34 } 35 else 36 { 37 cnt += c[ans]; 38 } 39 } 40 return ans + 1; 41 } 42 43 int main () 44 { 45 while ( scanf("%d", &n) != EOF ) 46 { 47 memset( c, 0, sizeof(c) ); 48 for ( int i = 1; i <= n; i++ ) 49 { 50 scanf("%d", a + i); 51 update( i, a[i] ); 52 } 53 scanf("%d", &m); 54 while ( m-- ) 55 { 56 char op[2]; 57 int x, y; 58 scanf("%s", op); 59 if ( op[0] == 'q' ) 60 { 61 scanf("%d", &x); 62 printf("%d ", kth(x)); 63 } 64 else 65 { 66 scanf("%d%d", &x, &y); 67 update( x, y - a[x] ); 68 a[x] = y; 69 } 70 } 71 } 72 return 0; 73 }
线段树：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 100001; 7 int a[N]; 8 int n, m; 9 10 struct Node 11 { 12 int l, r, sum; 13 } node[N << 2]; 14 15 void pushup( int i ) 16 { 17 node[i].sum = node[i << 1].sum + node[i << 1 | 1].sum; 18 } 19 20 void build( int i, int l, int r ) 21 { 22 node[i].l = l, node[i].r = r; 23 if ( l == r ) 24 { 25 node[i].sum = a[l]; 26 return ; 27 } 28 int mid = ( l + r ) >> 1; 29 build( i << 1, l, mid ); 30 build( i << 1 | 1, mid + 1, r ); 31 pushup(i); 32 } 33 34 void update( int i, int pos, int val ) 35 { 36 if ( node[i].l == pos && node[i].r == pos ) 37 { 38 node[i].sum = val; 39 return ; 40 } 41 int mid = ( node[i].l + node[i].r ) >> 1; 42 if ( pos <= mid ) 43 { 44 update( i << 1, pos, val ); 45 } 46 else 47 { 48 update( i << 1 | 1, pos, val ); 49 } 50 pushup(i); 51 } 52 53 int query( int i, int v ) 54 { 55 if ( node[i].l == node[i].r ) return node[i].l; 56 if ( node[i << 1].sum >= v ) return query( i << 1, v ); 57 return query( i << 1 | 1, v - node[i << 1].sum ); 58 } 59 60 int main () 61 { 62 while ( scanf("%d", &n) != EOF ) 63 { 64 for ( int i = 1; i <= n; i++ ) 65 { 66 scanf("%d", a + i); 67 } 68 build( 1, 1, n ); 69 scanf("%d", &m); 70 while ( m-- ) 71 { 72 char op[2]; 73 int x, y; 74 scanf("%s", op); 75 if ( op[0] == 'q' ) 76 { 77 scanf("%d", &x); 78 printf("%d ", query( 1, x )); 79 } 80 else 81 { 82 scanf("%d%d", &x, &y); 83 update( 1, x, y ); 84 } 85 } 86 } 87 return 0; 88 }