Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0

都选最优的，最多会多出一盒饼，然后减去最少的就行了

#include <bits\stdc++.h>using namespace std;#define pb push_back#define mp make_pairconst int maxn=1e5+9;vector< pair<long long ,long long> >aa,bb;bool cmp(pair<long long ,long long> a, pair<long long ,long long> b){    return a.first<b.first;}long long s[maxn],a[maxn],b[maxn];int main(){  //freopen("test.in","r",stdin);  int n;long long s1;long long len1,len2,ans,ans1,ans2,len;len=len1=len2=0;ans1=ans2=0;scanf("%d%I64d",&n,&s1);for(int i=1;i<=n;i++){    scanf("%I64d%I64d%I64d",&s[i],&a[i],&b[i]);    len+=s[i];    if(a[i]<b[i])    {        len1+=s[i];        ans1+=b[i]*s[i];        aa.pb(mp(b[i]-a[i],s[i]));    }    else    {        len2+=s[i];        ans2+=a[i]*s[i];        bb.pb(mp(a[i]-b[i],s[i]));    }}if((len/s1+(len%s1==0?0:1))>=((len2/s1+(len2%s1==0?0:1))+(len1/s1+(len1%s1==0?0:1))))//这个判断少加了点东西，终测挂了，挂我锁他了，结束之前就发现错误了    printf("%I64d\n",ans1+ans2);else{    len1=len1%s1;    len2=len2%s1;    len1=min(len1,s1-len2);    len2=min(len2,s1-len1);    long long tmp1,tmp2;    tmp1=tmp2=0;    long long sum1=0;    long long sum2=0;    sort(aa.begin(),aa.end(),cmp);    sort(bb.begin(),bb.end(),cmp);   for(int i=0;i<aa.size();i++)   {       if(tmp1+aa[i].second<len1)       {           tmp1+=aa[i].second;           sum1+=aa[i].second*aa[i].first;       }       else       {           sum1+=(len1-tmp1)*aa[i].first;           break;       }   }    for(int i=0;i<bb.size();i++)   {       if(tmp2+bb[i].second<len2)       {           tmp2+=bb[i].second;           sum2+=bb[i].second*bb[i].first;       }       else       {           sum2+=(len2-tmp2)*bb[i].first;           break;       }   }   // printf("%lld\n",ans1+ans2);   ans=ans1+ans2-min(sum1,sum2);   printf("%I64d\n",ans);}    return 0;}