标签：rest cas scan art medium scanf poj nis com
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 7891 Accepted: 2572 Special Judge
Description
Input
Output
Sample Input
5 1 -10 -5 0 5 10 3 10 2 -9 8 -7 6 -5 4 -3 2 -1 0 5 11 15 2 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 15 100 0 0
Sample Output
5 4 4 5 2 8 9 1 1 15 1 15 15 1 15
题意：求一个子区间权值和的绝对值与t相差最少。
题解：尺取法。尺取法维护的数列必须具有单调性，而此题中数值有正有负，不满足单调性。所以间接利用前缀和进行尺取法。
代码:
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=100005,inf=0x7fffffff;
struct node
{
int id,sum;
node(){}
node(int a,int b)
{id=a;sum=b;}
bool operator <(const node&n) const
{return sum<n.sum;}
}w[maxn];
int n;
void solve(int t);
int main()
{
int i,k,t,x;
while(~scanf("%d%d",&n,&k))
{
if(!n&&!k) break;
w[0]=node(0,0);
for(i=1;i<=n;i++)
{
scanf("%d",&x);
w[i]=node(i,w[i-1].sum+x);
}
sort(w,w+n+1);
while(k--)
{
scanf("%d",&t);
solve(t);
}
}
return 0;
}
void solve(int t)
{
int ansl,ansr,i,j,sum,mmin=inf,ans;
for(j=1,i=0;j<=n&&i<=n;)
{
sum=w[j].sum-w[i].sum;
if(abs(sum-t)<mmin)
{
ans=sum;
mmin=abs(sum-t);
ansl=w[i].id;
ansr=w[j].id;
}
if(sum>t) i++;
else if(sum<t) j++;
else break;
if(i==j) j++;
}
if(ansl>ansr) swap(ansl,ansr);
printf("%d %d %d\n",ans,ansl+1,ansr);
}
标签：rest cas scan art medium scanf poj nis com
原文地址：https://www.cnblogs.com/VividBinGo/p/11369983.html
