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a[i]=k*a[j]+m
m即是所求
从大到小枚举a[j]后,枚举倍数k
用到了stl里的low_bound()
#include"cstdio"
#include"cstring"
#include"algorithm"
#define MAXN 1000005
using namespace std;
int num[MAXN],n;
int get_ans(int x)
{ int temp=x,ans=0;
while(temp<num[n-1]){
temp+=x;
int pos=lower_bound(num,num+n,temp)-num;
if(!pos) continue;
else pos--;
ans=max(ans,num[pos]%x);
}
return ans;
}
int main()
{ while(scanf("%d",&n)!=EOF){
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
sort(num,num+n);
int ans=0;
for(int i=n-1;i>=0;i--){
if(ans>=num[i]) break;
if(num[i]==num[i+1]&&i!=n-1) continue;
ans=max(ans,get_ans(num[i]));
}
printf("%d\n",ans);
}
return 0;
}
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原文地址:http://www.cnblogs.com/luxiaoming/p/4705311.html