HDU 5496——Beauty of Sequence——————【考虑局部】
Beauty of Sequence
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 384 Accepted Submission(s): 168
Now you are given a sequence A of n integers {a1,a2,...,an}. You need find the summation of the beauty of all the sub-sequence of A. As the answer may be very large, print it modulo 109+7.
Note: In mathematics, a sub-sequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example {1,3,2} is a sub-sequence of {1,4,3,5,2,1}.
The first line contains an integer n (1≤n≤105), indicating the size of the sequence. The following line contains n integers a1,a2,...,an, denoting the sequence(1≤ai≤109).
The sum of values n for all the test cases does not exceed 2000000.
#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
const int mod=1e9+7;
const int maxn=1e5+200;
INT a[maxn];
INT pow2[maxn];
map<int,INT>coun;
int main(){
int T,n;
scanf("%d",&T);
pow2[0]=1;
for(int i=1;i<=maxn-100;i++){
pow2[i]=pow2[i-1]*2%mod;
}
while(T--){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%I64d",&a[i]);
}
coun.clear();
coun[0]=1;
INT ans=0;
for(int i=1;i<=n;i++){
INT pre=coun[a[i]];
INT tpre=pow2[i-1]-pre; //i-1
tpre=(tpre%mod+mod)%mod;
INT tmp=a[i]*tpre%mod*pow2[n-i]%mod;
ans=(ans+tmp)%mod;
pre=pow2[i-1]+pre;
coun[a[i]]=pre;
}
printf("%I64d\n",ans);
}
return 0;
}
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