HDU5857 Median[模拟]

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题意:给你一个一串已经排好序的数，再给你两个区间L1,R1,L2,R2

问这两个区间组成的新的一串数的中位数是多少。

思路:刚开始想可以分成三种情况包含，分离和相交

后来仔细一想，包含和相交可以通过交换R1,R2来合并,这样就只有两种情况了，模拟即可。

因为我在写新串偶数情况时先将两个数加起来，这样的话就暴int了，改成long long就过了，WA了无数发..

#include<bits/stdc++.h>using namespace std;long long a[100005];int main(){int t,n,m,l1,l2,r1,r2;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=n;++i){scanf("%lld",&a[i]);}while(m--){scanf("%d%d%d%d",&l1,&r1,&l2,&r2);if(l2<l1){swap(l1,l2);swap(r1,r2);}if(l2>=l1&&r2<=r1){swap(r1,r2);}long long len,L1,L2,L3;long long ans=0;if(l2>r1){len=r1-l1+1+r2-l2+1;L1=r1-l1+1;L2=r2-l2+1;if(len%2==0){if(len/2<=L1){ans+=a[l1+len/2-1];}else{ans+=a[l2+len/2-L1-1];}if(len/2+1<=L1){ans+=a[l1+len/2];}else{ans+=a[l2+len/2-L1];}printf("%.1lf\n",ans*1.0/2);}else{if(len/2+1<=L1){ans+=a[l1+len/2];}else{ans+=a[l2+len/2-L1];}printf("%.1lf\n",ans*1.0);}}else{len=(r1-l2+1)*2+l2-l1+r2-r1;L1=l2-l1;L2=r2-r1;L3=(r1-l2+1)*2;//cout<<len<<" "<<L1<<" "<<L2<<" "<<L3<<endl;if(len%2==0){if(len/2>L1&&len/2<=L1+L3){ans+=a[l2+(len/2-L1)/2-((len/2-L1)%2==0?1:0)];}else if(len/2<=L1){ans+=a[l1+len/2-1];}else if(len/2>L1+L3){ans+=a[r1+(len/2-L3-L1)];}if(len/2+1>L1&&len/2+1<=L1+L3){ans+=a[l2+(len/2+1-L1)/2-((len/2+1-L1)%2==0?1:0)];}else if(len/2+1<=L1){ans+=a[l1+len/2];}else if(len/2+1>L1+L3){ans+=a[r1+(len/2+1-L3-L1)];}printf("%.1lf\n",ans*1.0/2);}else{if(len/2+1>L1&&len/2+1<=L1+L3){ans+=a[l2+(len/2+1-L1)/2-((len/2+1-L1)%2==0?1:0)];}else if(len/2+1<=L1){ans+=a[l1+len/2];}else if(len/2+1>L1+L3){ans+=a[r1+(len/2+1-L3-L1)];}printf("%.1lf\n",ans*1.0);}}}}}

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