zoj 2029【二分查找】或【STL】

题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1029

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题目大意是说给出a数组，对于b中的每一个在a中找出最小范围使得ai<=b<a(i+1)，注意Print a blank line after each test case.这一句话！开始没看到结果WA了，杯具

自敲的二分

#include<iostream>using namespace std;#include<algorithm>int n,m;int find(int *a,int x){int l=0,r=n-1,mid;if(x<a[0] || a[n-1]<=x)return -1;while(l<=r){mid=(l+r)/2;if(a[mid]<=x && a[mid+1]>x)return mid;if(a[mid]<=x)l=mid;elser=mid;}return l;//这里不需要再return了，因为上面肯定可以找到这个区间}int main(){int a[10001];freopen("a.txt","r",stdin);while(cin>>n>>m){int i,j;for(i=0;i<n;i++)cin>>a[i];sort(a,a+n);for(i=0;i<m;i++){int b;cin>>b;j=find(a,b);if(j>=0)cout<<"["<<a[j]<<","<<a[j+1]<<")"<<endl;elsecout<<"no such interval/n";}cout<<"/n";}return 0;}

 

stl---lower_bound()

#include<iostream>using namespace std;#include<algorithm>int n,m;int find(int *a,int x){if(x<a[0] || a[n-1]<=x)return -1;return lower_bound(a,a+n,x)-a;}int main(){int a[10001];while(cin>>n>>m){int i,j;for(i=0;i<n;i++)cin>>a[i];sort(a,a+n);for(i=0;i<m;i++){int b;cin>>b;j=find(a,b);if(a[j]!=b)j=j-1;if(j>=0)cout<<"["<<a[j]<<","<<a[j+1]<<")"<<endl;elsecout<<"no such interval/n";}cout<<"/n";}return 0;}