NOIp2013火柴排队

看着纸质题做的，感觉挺水的一道题，结果居然过不了

思路和网上的正解非常相似，但是就是不知道哪里错了

我的做法复杂度更优秀，整体思路也基本一致，为什么不对呢？

可能还是没想明白吧，等我明天再想想。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 
 5 using namespace std;
 6 
 7 typedef long long ll;
 8 const ll mod = 99999997;
 9 
10 ll read(){
11     ll ans = 0;
12     char last = ' ',ch = getchar();
13     while(ch < '0'||ch > '9')last = ch,ch = getchar();
14     while('0' <= ch&&ch <= '9')ans = ans*10+ch-'0',ch = getchar();
15     if(last == '-')return -ans;return ans;
16 }
17 
18 struct node{
19     ll a,b;
20     bool operator <(const node& x)const{
21         return a < x.a;
22     }
23 }nbs[100010];
24 
25 ll a[100010];
26 ll merge_sort(int l,int r){
27     if(l >= r)return 0;
28     int mid = (l+r)>>1,p = l,q = mid+1;
29     int b[r-l+1],cnt = 0;
30     ll ans = 0;
31     ans += merge_sort(l,mid);
32     ans += merge_sort(mid+1,r);
33     while(p <= mid||q <= r){
34         if(p > mid||(q <= r&&a[q] < a[p])){
35             b[cnt++] = a[q++];
36             ans += mid-p+1;
37         }
38         else b[cnt++] = a[p++];
39     }
40     for(int i = l;i <= r;i++)a[i] = b[i-l];
41     return ans%mod;
42 }
43 
44 int n;
45 
46 int main(){
47     n = read();
48     for(int i = 1;i <= n;i++)nbs[i].a = read();
49     for(int i = 1;i <= n;i++)nbs[i].b = read();
50     sort(nbs+1,nbs+n+1);
51     for(int i = 1;i <= n;i++)a[i] = nbs[i].b;    
52     cout << merge_sort(1,n) << '\n';
53 return 0;
54 }

 

2020-10-30

来填坑了。。。

当年的我，太可爱了，连离散化都不知道。。。

我自己的ac代码：（现在会用树状数组了）

 1 int a[Maxn],b[Maxn],c[Maxn],t[Maxn];
 2 int n,ans;
 3 
 4 void add(int x,int y){
 5     while(x <= n){
 6         t[x] = (1ll*t[x]+y)%mod;
 7         x += x&-x;
 8     }
 9 }
10 
11 int ask(int x){
12     int ans = 0;
13     while(x){
14         ans += t[x];
15         x -= x&-x;
16     }
17     return ans;
18 }
19 
20 void work1(int a1[]){
21     rep(i,1,n)a1[i] = b[i] = read();
22     sort(a1+1,a1+n+1);
23     rep(i,1,n)b[i] = lower_bound(a1+1,a1+n+1,b[i])-a1;
24     rep(i,1,n)a1[b[i]] = i;
25 }
26 
27 void work2(int a1[]){
28     rep(i,1,n)a1[i] = b[i] = read();
29     sort(b+1,b+n+1);
30     rep(i,1,n)a1[i] = lower_bound(b+1,b+n+1,a1[i])-b;
31 }
32 
33 int main(){
34     n = read();
35     work1(a),work2(c);
36     per(i,n,1){
37         ans = (1ll*ans+ask(a[c[i]]-1))%mod;
38         add(a[c[i]],1);
39     }
40     cout << ans;
41 return 0;
42 }

 

posted @ 2019-09-06 23:58  TIH_HIT  阅读(199)  评论(0)  编辑  收藏  举报