BUPT2017 wintertraining(15) #5I
题意
输出序列A[1..n]的第n-1阶差分（一个整数）。
题解
观察可知答案就是
[sum_{i=0}^{n-1} {(-1)^{i}C_n^{i} A_{n-i}}
]
需要用大整数。
代码
Java代码
import java.io.*;
import java.math.*;
import java.util.*;
import java.text.*;
public class Main{
public static void main(String[] argc){
Scanner cin = new Scanner (new BufferedInputStream(System.in));
int T = cin.nextInt();
while(T>0){
--T;
int n = cin.nextInt();
BigInteger a, t, x;
a = BigInteger.ZERO;
t = BigInteger.ONE;
for (int i = 1; i <= n; ++i){
BigInteger b = BigInteger.ONE;
if (i >= 2){
t = t.multiply(BigInteger.valueOf(n-i+1));
t = t.divide(BigInteger.valueOf(i-1));
}
x = cin.nextBigInteger();
b = b.multiply(t);
b = b.multiply(x);
if (1 == (n - i) % 2) a = a.subtract(b);
else a = a.add(b);
}
System.out.println(a);
}
}
}
C++代码
#include <cstdio>
#include <cstring>
using namespace std;
#define N 3005
#define base 10000
struct Big{
int d[1000];
int k,l;
Big(){l=k=0;}
Big(int d){input(d);}
void input(int n){
memset(d,0,sizeof d);
k=0;
if(n==0) l=1;
else for(l=0;n;n/=base)d[l++]=n%base;
}
void output()const{
if(k&&l&&d[l-1])printf("-");
printf("%d",d[l-1]);
for(int i=l-2;i>=0;i--)printf("%04d",d[i]);
}
bool operator <(const Big &b)const{
int i=l-1,j=b.l-1;
if(i!=j)return i<j;
while(i>=0&&d[i]==b.d[j]){i--;j--;}
return d[i]<b.d[j];
}
Big operator * (const Big &b)const{
Big c;
memset(c.d,0,sizeof c.d);
for(int i=0;i<l;i++)
for(int j=0;j<b.l;j++){
c.d[i+j]+=d[i]*b.d[j];
if(c.d[i+j]>=base){
c.d[i+j+1]+=c.d[i+j]/base;
c.d[i+j]%=base;
}
}
c.l=l+b.l;
if(c.l>1&&!c.d[c.l-1])c.l--;
return c;
}
Big operator * (int b)const{
Big c;
c.input(b);
return *this*c;
}
Big operator / (int b)const{
Big c;
int i,k=0;
c.l=l;
for(i=l-1;i>=0;i--){
c.d[i]=(k+d[i])/b;
k=(k+d[i])%b*base;
}
while(c.l>1&&!c.d[c.l-1])c.l--;
return c;
}
Big operator + (const Big &b)const{
Big c=*this;
int k=0;
for(int i=0;i<b.l||i<l;i++){
if(i<b.l)c.d[i]+=b.d[i];
c.d[i]+=k;
k=0;
if(c.d[i]>=base){
c.d[i]-=base;
k=1;
}
}
if(b.l>l)c.l=b.l;
if(k)c.d[c.l++]=1;
return c;
}
Big operator - (const Big &b)const{
Big c=*this;
for(int i=0;i<b.l;i++){
c.d[i]-=b.d[i];
if(c.d[i]<0){
c.d[i]+=base;
c.d[i+1]--;
}
}
while(c.l>1&&!c.d[c.l-1])c.l--;
return c;
}
}a[N];
int T,n;
void add(Big &a,Big b){
if(a.k^b.k){
if(b<a) a=a-b;
else a=b-a,a.k=b.k;
}else a=a+b;
}
void sub(Big &a,Big b){
if(a.k^b.k) a=a+b;
else if(b<a) a=a-b;
else a=b-a,a.k^=1;
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i=1,d;i<=n;i++){
scanf("%d",&d);
a[i].input(d);
}
Big c(1);
int k=n%2;
Big ans=a[1];
if(!k)ans.k=1;
for(int i=1;i<n;i++){
c=c*(n-i)/i;
k^=1;
Big t=c*a[i+1];
if(k) add(ans,t);
else sub(ans,t);
}
ans.output();
puts("");
}
}