poj3378 树状数组 + 离散化 + 高精度

来源：互联网发布：苹果编程游戏编辑：程序博客网时间：2024/05/08 08:29

其实.,我没有AC,因为...高进度不会,所以没有搞.ok,这个是我直接一次后的代码.是为了以后再做这一题用的

#include <stdio.h>#include <string>#include <string.h>#include <queue>#include <stack>#include <map>#include <iostream>#include <stdlib.h>#include <math.h>#include <set>#include <algorithm>#define inf 0x3f3f3f3f#define mem0(x , y)  memset(x , y , sizeof(x))#define ll long long#define LL long long#define rep(x , y)   for(int x=0;x<y;x++)#define per(x , y)   for(int x=y-1;x>=0;x--)#define lowbit(x) (x & (-x))#define read(x) scanf("%d",&x)#define print(x) printf("%d\n" , x) ;#define middle int mid = (l + r)/2#define lson rt<<1 , l , mid#define rson rt<<1|1 , mid+1, r#define CASE int T ; int ca = 0 ; read(T) ; while(T--)#define RT tree[rt]#define FP freopen("1" , "r" , stdin)#define mem0(x , y) memset(x , y , sizeof(x))#define BIN(x) ll bin ; bin = (ll)1 << x ; ///2的x次方#define E Edge[i]#define lowbit(x) (x & (-x))const int MAXN = 500000 ;const int MAXV = 200000 ;const int MAXE = 200000 ;const int mod  = 1e8+7  ;using namespace std ;/****************前向星****************struct edge{    int s , e , w , next ;    bool operator < (const edge & e) const{        return w < e.w ;    }}Edge[MAXE] ;int head[MAXV] , hct = 0 ;void add_edge(int s,int e,int w){    Edge[hct].s = s ; Edge[hct].e =e ;Edge[hct].w = w ; Edge[hct].next = head[s] ;    head[s] = hct ++ ;}/*************************************/int BIT[1000000] ;const int MAXB = 1000000 ;void add(int x,int val){    x ++ ;    while(x < MAXB) BIT[x] += val , x += lowbit(x) ;}int query(int x){    int ans = 0 ;    while(x) ans += BIT[x] , x -= lowbit(x) ;    return ans ;}int hs[1000000] , ch = 0 ;int da[1000000] ;int dp[1000000][6] ;int find(int x){    return lower_bound(hs , hs+ch , x) - hs ;}int main(){    int n ;while(read(n)!=EOF){        ch = 0 ;        rep(i , n) read(da[i]) , hs[ch++] = da[i] ;        sort(hs , hs+ch) ; ch = unique(hs , hs+ch) - hs ;        for(int i=0;i<n;i++) dp[i][1] = 1 ;        for(int l=2;l<=5;l++){            mem0(BIT , 0) ;            for(int i=0;i<n;i++){                dp[i][l] = query(find(da[i])) ;                add(find(da[i]) , dp[i][l-1]) ;            }        }        int ans = 0 ;        for(int i=0;i<n;i++){            ans += dp[i][5] ;        }        printf("%d\n",ans) ;    }}

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