从给定的一维数组中找出49——168的个数。

#include<stdio.h>
int a[200] = {0,156,116,126,95,131,129,86,112,107,83,138,
113,122,121,92,131,139,109,113,49,147,123,112,
117,80,100,67,104,112,91,90,155,140,93,113,101,
121,145,113,131,136,163,131,68,89,111,104,101,
76,128,108,109,156,118,69,105,85,115,115,149,
99,109,138,133,146,109,136,91,78,88,120,79,
125,160,55,134,98,137,91,115,115,118,93,82,
102,137,151,96,85,114,111,85,109,97,105,102,
92,121,115,121,86,55,125,96,104,85,62,113,141,
107,86,133,127};
int b[200] = {0};
int main()
{
int n,i;
for(i = 0;i < 200;i ++) {

if(a[i] >= 49 && a[i] <= 168) b[a[i]] ++;
}
for(i = 49;i <= 168;i ++) {
if(b[i]) printf("%d %d\n",i,b[i]);
}
return 0;
}

#include <stdio.h>

int main()
{
int i;
int b[200]={0}; /*存放个数*/
for(i=0 ; i < len ; i++) /* len为数组长度*/
if(a[i]>=49&&a[i]<=168)