Series 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1067 Accepted Submission(s): 390
Problem Description
Let A be an integral series {A1, A2, . . . , An}.
The zero-order series of A is A itself.
The first-order series of A is {B1, B2, . . . , Bn-1},where Bi = Ai+1 - Ai.
The ith-order series of A is the first-order series of its (i - 1)th-order series (2<=i<=n - 1).
Obviously, the (n - 1)th-order series of A is a single integer. Given A, figure out that integer.
Input
The input consists of several test cases. The first line of input gives the number of test cases T (T<=10).
For each test case:
The first line contains a single integer n(1<=n<=3000), which denotes the length of series A.
The second line consists of n integers, describing A1, A2, . . . , An. (0<=Ai<=105)
Output
For each test case, output the required integer in a line.
Sample Input
2 3 1 2 3 4 1 5 7 2
Sample Output
0 -5
Author
BUPT
Source
2014 Multi-University Training Contest
6
由题意容易发现规律，即系数为二项式展开：C (n,n-1)=C(n,n)*n/1; C(n,n-2)=C(n,n-1)*(n-1)/2;
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"algorithm"
using namespace std;
#define LL __int64
#define N 300
#define M 100000000000
#define max(a,b) (a>b?a:b)
int a[3010];
LL c[N],ans[N],cc[N]; //数组大小，数组第一个元素存大数的长度，若长度为负数则代表大数小于零
void mul(LL*c,int x) //一个大数乘以一个整数
{
int i;
for(i=1;i<=c[0];i++)
c[i]=c[i]*x;
for ( i=1;i<=c[0];i++)
{
c[i+1]+=c[i]/M;
c[i]=c[i]%M;
}
while(c[c[0]+1])
{
c[0]++;
c[c[0]+1]=c[c[0]]/M;
c[c[0]]%=M;
}
}
void div(LL *c,int y) //一个大数除以一个整数
{
int i;
for ( i=c[0];i>1;i--)
{
c[i-1]+=(c[i]%y)*M;
c[i]/=y;
}
c[1]/=y;
while (c[c[0]]==0) c[0]--;
}
void add(LL *ans,LL *c) //一个大数加上一个大数，结果存入ans
{
int f=-1,i;
if (ans[0]<0)
{
if (-ans[0]>c[0]) f=1;
else if (-ans[0]<c[0]) f=0;
else
{
for (i=c[0];i>0;i--)
if (ans[i]>c[i])
{
f=1;
break;
}
else if (ans[i]<c[i])
{
f=0;
break;
}
if (i==0)
{
for(i=0;i<N;i++)
ans[i]=0;
ans[0]=1;
return;
}
}
}
else
{
int l=max(ans[0],c[0]);
for(i=1;i<=l;i++)
{
ans[i+1]+=(ans[i]+c[i])/M;
ans[i]=(ans[i]+c[i])%M;
}
while(ans[ans[0]+1])
{
ans[0]++;
ans[ans[0]+1]+=ans[ans[0]]/M;
ans[ans[0]]%=M;
}
return;
}
if(f!=-1)
{
if(f)
{
for(i=1;i<=-ans[0];i++)
{
ans[i]-=c[i];
if (ans[i]<0)
{
ans[i]+=M;
ans[i+1]-=1;
}
}
while(ans[-ans[0]]==0) ans[0]++;
}
else
{
for(i=1;i<=c[0];i++)
{
ans[i]=c[i]-ans[i];
if(ans[i]<0)
{
ans[i]+=M;
c[i+1]--;
}
}
ans[0]=c[0];
while(ans[ans[0]]==0) ans[0]--;
}
}
}
void sub(LL *ans,LL *c) //一个大数减去一个大数，结果存入ans
{
int i;
if (ans[0]<0)
{
int l=max(-ans[0],c[0]);
for (i=1;i<=l;i++)
{
ans[i+1]+=(ans[i]+c[i])/M;
ans[i]=(ans[i]+c[i])%M;
}
while (ans[-ans[0]+1])
{
ans[0]--;
ans[-ans[0]+1]+=ans[-ans[0]]/M;
ans[-ans[0]]%=M;
}
}
else
{
int f=0;
if (ans[0]>c[0]) f=1;
else if (ans[0]<c[0]) f=0;
else
{
for (i=ans[0];i>0;i--)
if (ans[i]>c[i])
{
f=1;
break;
}
else if (ans[i]<c[i])
{
f=0;
break;
}
if (i==0)
{
for(i=0;i<N;i++)
ans[i]=0;
ans[0]=1;
return;
}
}
if (f)
{
for (i=1;i<=ans[0];i++)
{
ans[i]-=c[i];
if (ans[i]<0)
{
ans[i+1]--;
ans[i]+=M;
}
}
while (ans[ans[0]]==0) ans[0]--;
}
else
{
for (i=1;i<=c[0];i++)
{
ans[i]=c[i]-ans[i];
if (ans[i]<0)
{
ans[i]+=M;
c[i+1]--;
}
}
ans[0]=-c[0];
while (ans[-ans[0]]==0) ans[0]++;
}
}
}
int main()
{
int T,n,i;
scanf("%d",&T);
while (T--)
{
scanf("%d",&n);
for (i=0;i<n;i++)
scanf("%d",&a[i]);
memset(c,0,sizeof(c));
memset(ans,0,sizeof(ans));
c[0]=c[1]=1;
ans[0]=0;
int flag=1;
if(n%2==0)
flag=-1;
for(i=0;i<n;i++)
{
if(i>0)
{
mul(c,n-i);
div(c,i);
}
memcpy(cc,c,sizeof(c));
mul(c,a[i]);
if(flag==1)
add(ans,c);
else
sub(ans,c);
flag*=-1;
memcpy(c,cc,sizeof(cc));
}
if (ans[0]<0)
{
printf("-");
ans[0]=-ans[0];
}
printf("%I64d",ans[ans[0]]);
for (i=ans[0]-1;i>0;i--)
printf("%011I64d",ans[i]);
printf("\n");
}
return 0;
}