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示例代码: #include <stdio.h> int main(void){ int t1 = 0 , t2 = 0 , l1 = 0 , l2 = 0 ; int v1 = 0 , v2 = 0 , t = 0 , s = 0 , l = 0 ; scanf("%d %d %d %d %d", 阅读全文
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思路: 1.当测试与被测试的芯片全部可以互相测试时,为好芯片; 示例代码: #include <stdio.h>#define N 20 int main(void){ int n = 0 ; int i = 0 , j = 0 , a = 0 , b = 0; int num[N][N] = {0 阅读全文
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示例代码: #include <stdio.h>#define N 1000000 int main(void){ int n = 0 , i = 0; char arr[N] , tmp[N] , c[2]; scanf("%d",&n); for (i = 0 ; i < n ; i ++) { 阅读全文
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思路: 先根据例子找出规律,列出递归要打印的数据; 示例代码: #include <stdio.h> int n = 0; void dg(int x){ int i = 0; if (x != 1) { printf("("); dg(x-1); printf(")"); } for (i = 1 阅读全文
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示例代码: #include <stdio.h>#include <string.h>#define N 10 char num[N] = {0} ; void yuyin(int n , int x){ if (x == 1 && (n == 9 || n == 5)) { printf("shi 阅读全文
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思路(贪心): 1.两边往中间逼近,步数少; 2.单个字符出现时只考虑移动到中间的步数,不做移动,因为这是最后进行,不影响结果; 示例代码: #include <stdio.h>#define N 8000 int main(void){ int n = 0 ; int i = 0 , j = 0 阅读全文
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解题思路: 1.先将可能的情况列出,根据分类确定计算的方式; 示例代码: #include <stdio.h>#define N 8 int main(void){ int i = 0 , j = 0 , k = 0; double num[N] = {0} , tmp[N] = {0}; doub 阅读全文
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示例代码: #include <stdio.h>#define N 30 int main(void){ int n = 0 , m = 0 , sum = 0; int i = 0 , j = 0 , k = 0 ,l = 0; int num[N][N] , tmp[N][N] , output 阅读全文
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代码示例: #include <stdio.h> int i = 0 ;int Primes(int a){ for (i = 2 ; i <= a/2 ; i ++) { if (a%i == 0) { return 0; } } return 1;} int main(void){ int a 阅读全文
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示例代码: #include <stdio.h>#define N 10 int main(void){ int i = 0 ; int len1 = 0 , len2 = 0 , flag = 0; char arr1[N] = {0} , arr2[N] = {0}; scanf("%s",&a 阅读全文