解析

巧夺天工

可以说是把two pointers玩明白了
把所有的可能减出来的结果升序排列一下
然后一个选取区间合法当且仅当这个区间含有1-n所有数的至少一个可能的差
然后就可以two pointers了

#include<bits/stdc++.h>
const int N=2e5+100;
const int mod=1e9+7;
#define ll long long
using namespace std;
inline ll read() {ll x(0),f(1);char c=getchar();while(!isdigit(c)) {if(c=='-')f=-1;c=getchar();}while(isdigit(c)) {x=(x<<1)+(x<<3)+c-'0';c=getchar();}return x*f;
}int n,m;
int a[8],b[N],tot;
struct node {int val,id;
} p[N*6];
bool cmp(node a,node b){return a.val<b.val;
}
int num[N],now;
inline void add(int pl){now+=++num[p[pl].id]==1;
}
int main() {for(int i=1; i<=6; i++) a[i]=read();sort(a+1,a+1+6);n=read();for(int i=1; i<=n; i++) b[i]=read();for(int i=1; i<=n; i++) {for(int j=1; j<=6; j++) {p[++tot]=(node){b[i]-a[j],i};}}sort(p+1,p+1+tot,cmp);int l=1,r=0,ans=2e9;while(1){while(now!=n&&r<=tot){now+=++num[p[++r].id]==1;}if(r>tot) break;ans=min(ans,p[r].val-p[l].val);now-=--num[p[l++].id]==0;}printf("%d\n",ans);
}
/*
1
281239
*/

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CF1413C Perform Easily(two pointers)

解析

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