761D Dasha and Very Difficult Problem
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题目链接:
http://codeforces.com/problemset/problem/761/D
题解:
题目大意:给出n(数字的个数),以及b数组的范围[l,r]。第一行表示的是数组a,第二行表示的数b数组的大小(并不是b数组),让你求出满足条件b数组。
这道题目,难点主要是要维护一个b数组的最小值,我们首先要对b数组按大小进行排序(便于以后的处理),然后就是要维护b数组的最小值了,这里我拿案例2做例子,排完序后,第一个值是8,b的最小下限是2,所以我们直接把2当成满足条件的第一数,然后是b-a的值为-6,所以,如果我们要满足条件的话,第二个应该为-5,然后将其加上第二个a的值4,看和是不是满足给出b的条件,如果不满足,那么就向上加,变为-4。。。。。。(注意最后还要将其按输入的顺序输出)
代码:
#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1e5+10;#define met(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3ftypedef long long ll;struct node{ int num1; int num2; int pos1; int pos2;}s[maxn];int n,l,r;int cmp1(node a,node b){ return a.pos1<b.pos1;}int cmp2(node a,node b){ return a.pos2<b.pos2;}int main(){ while(cin>>n>>l>>r) { for(int i=1;i<=n;i++) { cin>>s[i].num1; s[i].num2=0; s[i].pos2=i; } for(int i=1;i<=n;i++) cin>>s[i].pos1; sort(s+1,s+1+n,cmp1); int ans=l-s[1].num1; int step=l; s[1].num2=l; int len=2; int flag=0; for(int i=2;i<=n;i++) { ans++; while(ans+s[i].num1<l||ans+s[i].num1>r) { ans++; if(ans+s[i].num1>r) { flag=1; break; } } step=ans+s[i].num1; if(step>r||flag==1) { flag=1; break; } s[len++].num2=step; } if(flag) cout<<-1<<endl; else { sort(s+1,s+1+n,cmp2); for(int i=1;i<=n;i++) { if(i!=n) cout<<s[i].num2<<" "; else cout<<s[i].num2; } cout<<endl; } }}
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