UVa10474 Where is the Marble ? 有序数组二分找值,lower_bound / upper_bound
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题意:
给出n个数,先把各数从小到大排序,然后q次询问xi在数组中的位置,不存在则输出相应信息。
输入样例:
4 1
2
3
5
1
5
5 2
1
3
3
3
1
2
3
0 0
输出样例:
CASE# 1:
5 found at 4
CASE# 2:
2 not found
3 found at 3
//=======================================
数组从小到大有序。
lower_bound :查找“大于或者等于x”的第一个位置。
upper_bound:查找“大于x”的第一个位置。
#include <cstdio>#include <algorithm>using namespace std;const int maxn = 10000 + 5;int a[maxn];int main(){ int cas = 1, x; int n, q; while(~scanf("%d%d", &n, &q)) { if(n==0&&q==0) break; for(int i=1; i<=n; ++i) { scanf("%d", &a[i]); } sort(a+1, a+n+1); printf("CASE# %d:\n", cas++); while(q--){ scanf("%d",&x); int p = lower_bound(a+1, a+n+1, x) - a; if(a[p]==x){ printf("%d found at %d\n", x, p);} else printf("%d not found\n", x); } } return 0;}
0 0
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