（已解决）579B但答案仍旧错误terminate called after throwing an instance of ‘ std::bad_array

输入16进制16 ff 10
就会出现如上问题。应该是字符串问题。不过为什么？

#include<stdio.h>
#include<string.h>
char z[17]={'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
int main(){int a,b;char n[50]={};while(scanf("%d%s%d",&a,n,&b)!=EOF){int product=1;int len=strlen(n);//上面N数组长度已定义。int anum[len]={1,2,3},i,j;printf("len=%d\n",len);for(i=0;i<3;i++){j=15;while(j>=0){if(n[i]==z[j]){anum[i]=j;printf("%d\n",anum[i]);break;}else j--;}}//把字符串转换成整型数组  printf("n[2]=%d\n",n[2]);printf("anum[0]=%d\n",anum[0]);printf("anum[1]=%d\n",anum[1]);printf("anum[2]=%d\n",anum[2]);for(i=0;i<len;i++){printf("%d",anum[i]);}printf("\n");long long int y=0; while(len--){y+=anum[len]*product;product*=a;}//转换为10进制数 printf("y=%d\n",y);int bnum[50]={0};i=0,j=0;do{bnum[i++]=y%b;y/=b;}while(y!=0);//转换为b进制数字字符串len=i;for(i=0;i<len;i++){printf("%d ",bnum[i]);}printf("\n");char b[50];for(i=0;i<len;i++){b[i]=z[bnum[i]];}b[i]='\0';//转换为b进制字符串for(i=len-1;i>=0;i--){printf("%c",b[i]);} printf("\n");}return 0;
}

问题在于已定义n[50]，后面又求长度strlen，所以我写了个函数来求长度，但Oj仍然没法通过。

#include<stdio.h>
#include<string.h>
char z[17]={'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
int nlen(char n[]){int len=strlen(n);return len;
}
int main(){int a,b;char n[50]={};while(scanf("%d%s%d",&a,n,&b)!=EOF){int product=1;int len=nlen(n);int anum[len]={},i,j;//printf("len=%d\n",len);for(i=0;i<3;i++){j=15;while(j>=0){if(n[i]==z[j]){anum[i]=j;//printf("%d\n",anum[i]);break;}else j--;}}//把字符串转换成整型数组  /*printf("n[2]=%d\n",n[2]);printf("anum[0]=%d\n",anum[0]);printf("anum[1]=%d\n",anum[1]);printf("anum[2]=%d\n",anum[2]);for(i=0;i<len;i++){printf("%d",anum[i]);}printf("\n");*/long long int y=0; while(len--){y+=anum[len]*product;product*=a;}//转换为10进制数 //printf("y=%d\n",y);int bnum[50]={0};i=0,j=0;do{bnum[i++]=y%b;y/=b;}while(y!=0);//转换为b进制数字字符串len=i;/*for(i=0;i<len;i++){printf("%d ",bnum[i]);}printf("\n");*/char b[50];for(i=0;i<len;i++){b[i]=z[bnum[i]];}b[i]='\0';//转换为b进制字符串for(i=len-1;i>=0;i--){printf("%c",b[i]);} printf("\n");}return 0;
}

正确代码：memcoy0
人家直接‘9’-‘0’，‘f’-‘a’+10;这样来计算值。和我方法不一样。

#include <cstdio>
#include <cstring>int CharToOct(int a, char n[]) {  //按进制a将n字符串(可表示2-16进制)转换为10进制数int sum = 0, product = 1;for (int i = strlen(n) - 1; i >= 0; i--) {if (n[i] <= '9') sum += (n[i] - '0') * product;else if (n[i] <= 'F') sum += (n[i] - 'A' + 10) * product; //大写字母符号else if (n[i] <= 'f') sum += (n[i] - 'a' + 10) * product; //小写字母符号product *= a;}return sum;
}void OctToChar(int temp, int b, char r[]) { //将10进制数按b进制转换成r字符串int i = 0;do {int k = temp % b;if (k <= 9) r[i++] = '0' + k; //十进制符号else r[i++] = 'A' + k - 10;   //用大写字母表示大于9的数字temp /= b;} while (temp != 0);r[i] = '\0';  //必须添加结束符, 不然strlen无法正确判别长度
}int main() {int a, b; // 2-16char n[100];while (scanf("%d%s%d", &a, n, &b) != EOF) {int temp = CharToOct(a, n);if (b == 10) {printf("%d\n", temp);continue;}char r[100];OctToChar(temp, b, r);for (int j = strlen(r) - 1; j >= 0; j--) printf("%c", r[j]);printf("\n");}return 0;
}

（已解决）579B但答案仍旧错误terminate called after throwing an instance of ‘ std::bad_array_length‘，相关推荐

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（已解决）579B但答案仍旧错误terminate called after throwing an instance of ‘ std::bad_array_length‘，

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