hdu5147 (sequence 2) 树状数组
1 #include <iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #include<vector> 6 #include<map> 7 using namespace std; 8 #define INF 0x7fffffff 9 #define MOD 1000000007 10 11 12 #define N 100010 13 #define MAX 1000100 14 15 int C[MAX], S[MAX], b[N],d[N]; 16 long long f[N]; 17 long long total[N], ans; 18 int num[N], T, s, t, i, j; 19 20 int Lowbit(int x){ 21 return x&(x^(x-1)); 22 } 23 24 void add(int pos,int num,int *P) { 25 while(pos <= MAX) { 26 P[pos] += num; 27 pos += Lowbit(pos); 28 } 29 } 30 31 int Sum(int end,int *P) { 32 int cnt = 0; 33 while(end > 0) { 34 cnt += P[end]; 35 end -= Lowbit(end); 36 } 37 return cnt; 38 } 39 40 void init(){ 41 total[0] = 0; 42 for(i = 1; i < N; ++i){ 43 total[i] = total[i-1] + i; 44 } 45 } 46 47 int main() { 48 //init(); 49 int t; 50 cin>>t; 51 52 while(t--) { 53 54 scanf("%d",&T); 55 56 memset(C,0,sizeof(C)); 57 memset(S,0,sizeof(S)); 58 memset(b,0,sizeof b); 59 memset( d,0,sizeof d); 60 memset(f,0,sizeof f); 61 memset(num,0,sizeof num); 62 //memset(num,0,sizeof(num)); 63 //memset(b,0,sizeof(b)); 64 //ans = 0; 65 for(j = 0; j < T; j ++) {//因为第一个数前面比它小的数没有，所以j要从0开始 66 scanf("%d",&num[j]); 67 add(num[j]+1,1,C); 68 b[j] = Sum(num[j], C);//Sum(num[j],C)求的就是小于s的个数，j - Sum(num[j],C)就是前j个数中大于num[j]的个数 69 //printf("%d ",b[j]); 70 } 71 //printf("\n"); 72 //ans = 0; 73 for(j = T-1; j > -1; --j){//反过来求第j个数右边中大于它的数的个数。 74 add(num[j]+1 ,1, S); 75 d[j] =T-1-j-Sum(num[j] ,S);//Sum(num[j],S)求的就是小于num[j]的个数 76 //b[j] -= Sum(num[j]+1,S) - Sum(num[j],S)-1; 77 //printf("%d ",b[j]); 78 //ans += total[b[j]]; 79 } 80 f[T-1]=0; 81 for(int j=T-2;j>=0;j--) 82 f[j]=d[j+1]+f[j+1]; 83 long long int res=0; 84 for(int j=0;j<T;j++) 85 res+=b[j]*f[j]; 86 cout<<res<<endl; 87 88 } 89 return 0; 90 } 91 92 代码君
理解有问题的话，先看树状数组求逆序数。
