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[100005];
ll b[100005];
ll suma[100005];
ll sumb[100005];
ll ans;
ll gg[100005];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        suma[0]=0;
        sumb[0]=0;
        a[0]=0;
        b[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            suma[i]=(suma[i-1]+a[i])%mod;
        }
            for(int i=1;i<=n;i++)
        {
            scanf("%d",&b[i]);
            sumb[i]=(sumb[i-1]+b[i])%mod;
        }
        ans=0;
        for(int i=1;i<=n;i++)
        {
          ans=((a[i]*b[i]%mod+a[i]*sumb[i-1]%mod+b[i]*suma[i-1]%mod))%mod;
          gg[i]=ans;
        }
        printf("%lld",gg[1]);
        for(int i=2;i<=n;i++)
         printf(" %lld",gg[i]);
         cout<<endl; 
    }
    return 0;
}