HDU 1592 Half of and a Half（高精度）
发布时间：2021-01-25 18:57:56 所属栏目：大数据 来源：网络整理
导读：Half of and a Half Time Limit: 1000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1139????Accepted Submission(s): 513 Problem Description Gardon bought many many chocolates from the A Chocolate Mar
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Half of and a HalfTime Limit: 1000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1139????Accepted Submission(s): 513 Problem Description Gardon bought many many chocolates from the A Chocolate Market (ACM). When he was on the way to meet Angel,he met Speakless by accident. ? "Ah,so many delicious chocolates! I'll get half of them and a half!" Speakless said. Gardon went on his way,but soon he met YZG1984 by accident.... "Ah,so many delicious chocolates! I'll get half of them and a half!" YZG1984 said. Gardon went on his way,but soon he met Doramon by accident.... "Ah,so many delicious chocolates! I'll get half of them and a half!" Doramon said. Gardon went on his way,but soon he met JGShining by accident.... "Ah,so many delicious chocolates! I'll get half of them and a half!" JGShining said. . . . After had had met N people,Gardon finally met Angel. He gave her half of the rest and a half,then Gardon have none for himself. Could you tell how many chocolates did he bought from ACM? ? Input Input contains many test cases. Each case have a integer N,represents the number of people Gardon met except Angel. N will never exceed 1000; ? Output For every N inputed,tell how many chocolates Gardon had at first. ? Sample Input 2? Sample Output 7? Author DYGG ? Source HDU “Valentines Day” Open Programming Contest 2007-02-14 ? /* HDU 1592靠，竟然是大数，刚才以为1000只是个什么小范围，2的1000+1次方啊 2^(n+1)-1，用大数怎么实现啊 高精度，貌似，好吧，，， */ #include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1005][105];
void System()
{
a[0][1] = 1;
for(int i = 1;i < 1005;i++)
{
for(int j = 1;j < 105;j++)
{
if(j == 1)
a[i][j] += a[i-1][j]*2 + 1;
else
a[i][j] += a[i-1][j]*2;
a[i][j+1] += a[i][j] / 10000;
a[i][j] = a[i][j] % 10000;
}
}
}
int main()
{
System();//高精度
int n;
while(~scanf("%d",&n))
{
int i,j;
for(i = 100;i >= 1;i--)
{
if(a[n][i])
{
printf("%d",a[n][i]);
break;
}
}
for(int j = i-1;j >= 1;j--)
printf("%04d",a[n][j]);
printf("n");
}
return 0;
}
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