1 条题解
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#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<cstdlib> #include<algorithm> #define ll long long #define N 25003 using namespace std; inline int read(int &ret) { ret=0;char ch=getchar(); while(ch<'0'||ch>'9') ch=getchar(); while('0'<=ch&&ch<='9') { ret=ret*10-'0'+ch; ch=getchar(); } } int n,m,a[N],b[N]; int main() { //freopen("spicychicken.in","r",stdin); //freopen("spicychicken.out","w",stdout); read(a[0]);read(b[0]);read(n);read(m); for(int i=n;i;--i)read(a[i]); sort(a,a+n+1,greater<int>()); for(int i=0;i<=n;++i)a[i]=a[i]-a[i+1]; sort(a,a+n+1); for(int i=m;i;--i)read(b[i]); sort(b,b+m+1,greater<int>()); for(int i=0;i<=m;++i)b[i]=b[i]-b[i+1]; sort(b,b+m+1); int p=1,q=1; ll ans=(ll)a[0]*m+(ll)b[0]*n; while(p<=n&&q<=m) if(a[p]<b[q])ans+=(ll)a[p++]*(m-q+1); else ans+=(ll)b[q++]*(n-p+1); printf("%lld\n",ans); return 0; }
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- 递交数
- 7
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- 通过率
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