Problem Description给出一段连续的内存,假设它的值是x,请输出x*16+7.Input多组测试数据,每组数据一行,仅由0和1组成.每行长度不超过60,并且不含前导0.Output对每组数据输出一行,表示x*16+7.注意：输出

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ProblemDescription给出一段连续的内存,假设它的值是x,请输出x*16+7.Input多组测试数据,每组数据一行,仅由0和1组成.每行长度不超过60,并且不含前导0.Output对每组

Problem Description给出一段连续的内存,假设它的值是x,请输出x*16+7.Input多组测试数据,每组数据一行,仅由0和1组成.每行长度不超过60,并且不含前导0.Output对每组数据输出一行,表示x*16+7.注意：输出
Problem Description
给出一段连续的内存,假设它的值是x,请输出x*16+7.
Input
多组测试数据,每组数据一行,仅由0和1组成.
每行长度不超过60,并且不含前导0.
Output
对每组数据输出一行,表示x*16+7.
注意：输出必须是转化为二进制后的结果.
Sample Input
10
Sample Output
100111
Hint
(10)2=2,2*16+7=39,(39)2=100111.

Problem Description给出一段连续的内存,假设它的值是x,请输出x*16+7.Input多组测试数据,每组数据一行,仅由0和1组成.每行长度不超过60,并且不含前导0.Output对每组数据输出一行,表示x*16+7.注意：输出
#include<stdio.h>
#include<string.h>
int main()
{
char str[100];
while(scanf("%s",str)!=EOF){
printf("%s",str);
printf("0111\n");
}
return 0;
}

思路：x*16等于x*2^4,相当于x左移4位,又x加7,及x后面加0111,输出即可.

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