G - For Fans of Statistics
Time Limit:1000MS
Memory Limit:65536KB
64bit IO Format:%I64d & %I64u Submit Status
Description
Have you ever thought about how many people are transported by trams every year in a city with a ten-million population where one in three citizens uses tram twice a day? Assume that there are
n cities with trams on the planet Earth. Statisticians counted for each of them the number of people transported by trams during last year. They compiled a table, in which cities were sorted alphabetically. Since city names were inessential for statistics, they were later replaced by numbers from 1 to
n. A search engine that works with these data must be able to answer quickly a query of the following type: is there among the cities with numbers from
l to
r such that the trams of this city transported exactly
x people during last year. You must implement this module of the system.
Input
The first line contains the integer
n, 0 <
n < 70000. The second line contains statistic data in the form of a list of integers separated with a space. In this list, the
ith number is the number of people transported by trams of the
ith city during last year. All numbers in the list are positive and do not exceed 10
9 ? 1. In the third line, the number of queries
q is given,
0 <
q < 70000.
The next
q lines contain the queries. Each of them is a triple of integers
l,
r, and
x separated with a space;
1 ≤
l ≤
r ≤
n
; 0 <
x < 10
9.
Output
Output a string of length
q in which the
ith symbol is “1” if the answer to the
ith query is affirmative, and “0” otherwise.
Sample Input
| input |
output |
5
1234567 666666 3141593 666666 4343434
5
1 5 3141593
1 5 578202
2 4 666666
4 4 7135610
1 1 1234567
|
10101
|
没发现和stl有什么关系。。。。。用二分过的，先以值得大小二分，如果值相同的话，加一组判断，以当前的id与给出的l和r进行二分，看最后能不能得出符合条件的值
#include
#include
#include
using namespace std; struct node{ int id , k ; } p[80000]; bool cmp(node a,node b) { return a.k
= l && p[mid].id <= r ) { printf("1"); return ; } else if( p[mid].k < x || ( p[mid].k == x && p[mid].id < l ) ) low = mid+1 ; else top = mid - 1 ; } printf("0"); return ; } int main() { int i , n , m , l , r , x ; while(scanf("%d", &n)!=EOF) { for(i = 0 ; i < n ; i++) { scanf("%d", &p[i].k); p[i].id = i+1 ; } sort(p,p+n,cmp); scanf("%d", &m); while(m--) { scanf("%d %d %d", &l, &r, &x); f(l,r,x,n); } printf("\n"); } }