2014 Multi-University Training Contest 4

官方题解：http://blog.sina.com.cn/s/blog_6bddecdc0102uytw.html

每一步都要mod，浪费了220分钟。

 1 #include<cstdio>
 2 #include<cstring>
 3 #define mt(a,b) memset(a,b,sizeof(a))
 4 typedef __int64 LL;
 5 const int mod=1000000007;
 6 const int M=1024;
 7 int a[M];
 8 LL dpxor[M][M],dpand[M][M];
 9 int main(){
10     int t,n;
11     while(~scanf("%d",&t)){
12         while(t--){
13             scanf("%d",&n);
14             for(int i=1;i<=n;i++){
15                 scanf("%d",&a[i]);
16             }
17             mt(dpxor,0);
18             mt(dpand,0);
19             for(int i=1;i<=n;i++){
20                 dpxor[i][a[i]]=1;
21                 for(int j=0;j<M;j++){
22                     dpxor[i][j^a[i]]+=dpxor[i-1][j];
23                     dpxor[i][j^a[i]]%=mod;
24                     dpxor[i][j]+=dpxor[i-1][j];
25                     dpxor[i][j]%=mod;
26                 }
27             }
28             for(int i=n;i>=1;i--){
29                 dpand[i][a[i]]=1;
30                 for(int j=0;j<M;j++){
31                     dpand[i][j&a[i]]+=dpand[i+1][j];
32                     dpand[i][j&a[i]]%=mod;
33                     dpand[i][j]+=dpand[i+1][j];
34                     dpand[i][j]%=mod;
35                 }
36             }
37             LL ans=0;
38             for(int i=1;i<=n;i++){
39                 for(int j=0;j<M;j++){
40                     ans+=((dpxor[i][j]+mod-dpxor[i-1][j])%mod)*dpand[i+1][j];
41                     ans%=mod;
42                 }
43             }
44             printf("%I64d\n",ans);
45         }
46     }
47     return 0;
48 }

View Code