HDU2688-Rotate
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For example initial sequence is 1 2 3 4 5.
If changing format is [1 3], than the sequence will be 1 3 4 2 5 because the first sequence is base from 0.
InputThe input contains multiple test cases.
Each case first given a integer n standing the length of integer sequence (2<=n<=3000000)
Second a line with n integers standing F[i](0<F[i]<=10000)
Third a line with one integer m (m < 10000)
Than m lines quiry, first give the type of quiry. A character C, if C is ‘R’ than give the changing format, if C equal to ‘Q’, just put the numbers of satisfy pairs.
OutputOutput just according to said.Sample Input
5 1 2 3 4 5 3 Q R 1 3 Q
Sample Output
10 8
题解
这道题我们可以用树状数组求出原数组的答案
因为abs(E-S)<=1000,m<10000,所以我们可以枚举每次的区间,因为翻转每次是S+1~E往前移一位,第S个到E
所以我们先把第S位存下来,每次和后面的判断一下大小就可以了
当输入是Q的时候就直接输出就可以了
1 #include<algorithm> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #define ll long long 6 #define M 10005 7 #define N 3000005 8 using namespace std; 9 int n,m,x,y; 10 ll ans; 11 int a[N]; 12 ll tr[M]; 13 char s[10]; 14 int lowbit(int x){ return x&(-x); } 15 void add(int x){ 16 while (x<=M){ 17 tr[x]++; 18 x+=lowbit(x); 19 } 20 } 21 int query(int x){ 22 int s=0; 23 while (x>0){ 24 s+=tr[x]; 25 x-=lowbit(x); 26 } 27 return s; 28 } 29 int main(){ 30 while (~scanf("%d",&n)){ 31 memset(tr,0,sizeof(tr)); ans=0; 32 for (int i=0;i<n;i++){ 33 scanf("%d",&a[i]); 34 add(a[i]); 35 ans+=query(a[i]-1); 36 } 37 scanf("%d",&m); 38 for (int i=1;i<=m;i++){ 39 scanf("%s",s); 40 if (s[0]==‘Q‘) printf("%lld\n",ans); 41 else{ 42 scanf("%d%d",&x,&y); 43 int s=a[x]; 44 for (int i=x;i<y;i++){ 45 a[i]=a[i+1]; 46 if (s<a[i]) ans--; else 47 if (s>a[i]) ans++; 48 } 49 a[y]=s; 50 } 51 } 52 } 53 return 0; 54 }
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