2014 UESTC Training for Data Structures—— E - 休生伤杜景死惊开
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解题思路:树状数组。假设x[i]是数值为i的数的个数,所以求的比q[i]小的数就有x[i-1]+x[i-2]+…..+x[2]+x[1],于是就可以用树状数组来写了,从头到尾扫一遍,对每个数都算一下前面比他小的数的个数即x[q[i]-1]+x[q[i]-2]+…….x[2]+x[1],然后从后往前扫一遍,对每个数计算后面比他小的数的数量。然后相同位置计算得到的两个数相乘,再求和就可以了。详见代码:
Code:
#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define MAX 50010int Lefts[MAX],Rights[MAX];//Lefts[i]用来记录从左到i小于i的个数,Rights[i]记录i到Nint Arrays[MAX],Maps[MAX];//Arrays[i]用来接收输入的数据,Maps[i]用来记录数值i出现的次数int maxn,N;//maxn记录输入数据的最大值,N为输入个数void Solve(int j){ while(j<=maxn) { Maps[j] +=1;//为大于或等于j的数值加1 j +=(j&-j);//然后j增加 }}int Counts(int j){ int ans = 0; while(j>0) { ans +=Maps[j];//累加满足条件的j的个数 j -=(j&-j);//因为j在提前处理时增加的是(j&-j),所以从j往后扫每次减少(j&-j) } return ans;}int main(){ //freopen("input.txt","r",stdin); while(~scanf("%d",&N)) { maxn = 0; memset(Maps,0,sizeof(Maps)); for(int i = 1; i<=N; i++) { scanf("%d",&Arrays[i]); if(Arrays[i]>maxn) maxn = Arrays[i];//找出最大值 } for(int i = 1; i<=N; i++)//从左往右处理 { Solve(Arrays[i]); Lefts[i] = Counts(Arrays[i]-1); } memset(Maps,0,sizeof(Maps)); for(int i = N; i>0; i--)//从右往左处理 { Solve(Arrays[i]); Rights[i] = Counts(Arrays[i]-1); } long long Sum = 0;//此处为防止累加越界故用long long for(int j = 1; j<=N; j++) Sum +=Lefts[j]*Rights[j];//将每个位置的左右两边个数相乘再进行累加 printf("%lld\n",Sum); } return 0;}
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- 2014 UESTC Training for Data Structures—— E - 休生伤杜景死惊开
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