ACdream 1029 前缀和

Multiplication

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

If C=A⋄B, then

C[k]=∑max(i,j)=kA[i]⋅B[j]mod(109+7)

.

Work out sequence C=A⋄Bfor given sequence A and B.

Input

The first line contains a integer n, which denotes the length of sequence A,B.

The second line contains nn integers a1,a2,…,ana1,a2,…,an , which denote sequence A.

The thrid line contains nn integers b1,b2,…,bnb1,b2,…,bn , which denote sequenceB.

(1≤n≤105,0≤ai,bi≤109)

Output

nn integers, which denotes sequence C.

Sample Input

2
1 2
3 4

Sample Output

3 18 

Source

ftiasch

Manager

nanae

 

题意：  长度为n的a，b序列   求C[k]=∑max(i,j)A[i]⋅B[j]mod(109+7)

 

题解：  max（i，j） 可以看出 相乘的两个数中一定有a[n]或者b[n]

          前缀和处理

 

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #define ll long long
 #define pi acos(-1.0)
 #define mod 1000000007
 using namespace std;
 int n;
 ll a[];
 ll b[];
 ll suma[];
 ll sumb[];
 ll ans;
 ll gg[];
 int main()
 {
     while(scanf("%d",&n)!=EOF)
     {
         suma[]=;
         sumb[]=;
         a[]=;
         b[]=;
         for(int i=;i<=n;i++)
         {
             scanf("%d",&a[i]);
             suma[i]=(suma[i-]+a[i])%mod;
         }
             for(int i=;i<=n;i++)
         {
             scanf("%d",&b[i]);
             sumb[i]=(sumb[i-]+b[i])%mod;
         }
         ans=;
         for(int i=;i<=n;i++)
         {
           ans=((a[i]*b[i]%mod+a[i]*sumb[i-]%mod+b[i]*suma[i-]%mod))%mod;
           gg[i]=ans;
         }
         printf("%lld",gg[]);
         for(int i=;i<=n;i++)
          printf(" %lld",gg[i]);
          cout<<endl;
     }
     return ;
 }