标签：map bool cst ssi 连续 log which not research
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：找连续的数使得和的绝对值和给定数差最小
题解：想了很久没想到，****因为尺取法必须单调数列才行，而且是求连续和****，先求前缀和，给其排序，再进行尺取
因为s不能等于t，当s==t时，t要++；sum>k时，向后移动s++，反之t++；
#include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define pi acos(-1) #define ll long long #define mod 1000000007 using namespace std; const int N=100000+5,maxn=10000+5,inf=0x3f3f3f3f; struct edge{ int v,id; }a[N]; int n; bool comp(const edge &a,const edge &b) { return a.v<b.v; } void solve(int x) { int s=0,t=1,ss,tt,sum,ans,minn=inf; while(s<=n&&t<=n&&minn!=0){ sum=a[t].v-a[s].v; if(abs(sum-x)<minn) { minn=abs(sum-x); ans=sum; ss=a[s].id; tt=a[t].id; } if(sum>x)s++; else if(sum<x)t++; else break; if(t==s)t++; } if(ss>tt)swap(ss,tt); cout<<ans<<" "<<ss+1<<" "<<tt<<endl; } int main() { ios::sync_with_stdio(false); cin.tie(0); int k,s; while(cin>>n>>k,n||k){ a[0].v=a[0].id=0; for(int i=1;i<=n;i++) { int p; cin>>p; a[i].v=a[i-1].v+p; a[i].id=i; } sort(a,a+n+1,comp); while(k--){ cin>>s; solve(s); } } return 0; }
标签：map bool cst ssi 连续 log which not research
原文地址：http://www.cnblogs.com/acjiumeng/p/6754683.html
