Submission #1370904

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Source Code Expand

#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int N,K;
ll A[101010],B[101010];
ll S[101010];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>K;
	FOR(i,N) {
		cin>>A[i]>>B[i];
		if(B[i]==1 && A[i]>K/2) return _P("-1\n");
		if(B[i]==2) return;
		S[i+1]=S[i]+A[i];
	}
	ll mi=1LL<<60;
	for(i=N;i>=0;i--) {
		if(S[N]-S[i]>K/2) continue;
		
		ll cur=S[i];
		ll wait=0;
		for(j=i;j>0;j--) {
			ll L=cur%K,R=(cur+A[j-1])%K;
			if(R<L) R+=L;
			ll x=S[j-1]%K,y=S[j]%K;
			if(y<x) y+=x;
			
			if(R<=x || L>=y) {
				cur+=A[j-1];
			}
			else {
				cur+=(y-L)+A[j-1];
				wait+=(y-L);
			}
		}
		mi=min(mi,wait);
		
	}
	cout<<2*S[N]+mi<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}

Submission Info

Judge Result