标签：style blog class code java c
二分查找的一点思考
二分查找算法实现
#define LOCAL #include<cstdio> #include<cstring> #include<algorithm> int const MAX_N=2<<10; int const MAX_M=100; int a[MAX_N],i,n,lb,ub,k; void solve() { for(i=0;i<n;i++) { scanf("%d",&a[i]); } scanf("%d",&k); //排序 std::sort(a,a+n); //搜索 lb=-1;ub=n; while(ub-lb>1) { int mid=(ub+lb)/2; if(a[mid]>=k) { ub=mid; }else{ lb=mid; } } for(i=0;i<n;i++) { printf("a[%d]=%d\t",i,a[i]); } printf("\n%d\n",ub); } int main() { #ifdef LOCAL freopen("3.2.in","r",stdin); freopen("3.2.out","w",stdout); #endif memset(a,0,sizeof(a)); while(~scanf("%d",&n)) { solve(); } return 0; }
另一种版本：
#define LOCAL #include<cstdio> #include<cstring> #include<algorithm> int const MAX_N=2<<10; int const MAX_M=100; int a[MAX_N],i,n,lb,ub,k; void solve() { for(i=0;i<n;i++) { scanf("%d",&a[i]); } scanf("%d",&k); //排序 std::sort(a,a+n); lb=-1;ub=n; //搜索 for(i=0;i<MAX_M;i++) { int mid=(ub+lb)/2; if(a[mid]>=k) { ub=mid; }else{ lb=mid; } } for(i=0;i<n;i++) { printf("%d ",a[i]); } printf("\n%d\n",ub); } int main() { #ifdef LOCAL freopen("3.2.in","r",stdin); freopen("3.2.out","w",stdout); #endif memset(a,0,sizeof(a)); while(~scanf("%d",&n)) { solve(); } return 0; }
一点思考：
#define LOCAL #include<cstdio> #include<cstring> #include<algorithm> int const MAX_N=2<<10; int const MAX_M=100; void solve() { int i; //就是说1经过MAX_N次循环，能从1变到1267650600228229400000000000000.000000 double first=1; for(i=0;i<MAX_M;i++) { first=first*2; } printf("%lf\n",first); //试想一根1267650600228229400000000000000.000000长的绳子，每次截半，100次之后就变成1了 //由此对付一个足够的的数组，采用二分查找MAX_N次已经是绰绰有余了 } int main() { #ifdef LOCAL freopen("3.2.in","r",stdin); freopen("3.2.out","w",stdout); #endif solve(); return 0; }
标签：style blog class code java c
原文地址：http://www.cnblogs.com/jianfengyun/p/3724182.html
