weiwanshu的专栏

#includeint main(void){ int n, i, m, a[10000]={0}; scanf("%d", &n); for(i = 1; i <= n; i++){ scanf("%d", &a[i]); } scanf("%d", &m); for( i = 1; i <= n; i++){ //之前出现一个错误，写成了i = 0，结果过了9

这道题很简单，测试数据也很水，按理说我的程序遇到大一点的数据就过不去了，可是后台只有两组数据，最大的只到2000+，数组开大一点不就好啦。。。于是一开始，开了一个巨大巨大的数组，这么蠢的方法真是不好意思

左边图是输入3时的结果，右边是输入1时的结果，之前曾经想要不要用规划，查了一下发现没有必要，把图形分成四块，每块又分成三个部分，开个字符数组赋值输出就可以了，很简单

这可以看成是一道全排列的问题思路：1、全排列出所有可能的数列（全排列问题留到稍后说）2、对数列进行分析判断

#includeint main(void){ int n; scanf("%d", &n); int a[100][100] = {1}, i, j; for( i = 0; i <= n-1; i++){ a[i][0] = 1; printf("%d",a[i][0]); for( j = 1; j <= i; j++){ a[i][j] = a[i-1][j-1

#includeint main(void){ long n; int a[16] = {0}, i, j = 0; scanf("%ld", &n); for( i = 0; n >= 1; i++){ a[i] = n % 16; n = n / 16; j = i; } for( j; j >= 0; j--){ printf("%X", a[j]); }

#includeint main(void){ int a, b; for( a = 1; a <= 9; a++){ for( b = 0; b <= 9; b++){ printf("%d%d%d%d\n", a, b, b, a); } } return 0;}

#include#include#includeint main(void){ long long n = 0; char a[9]; int m; scanf("%d", &m); getchar(); while( m--){ gets(a); int s; s = strlen(a); int i; for( i = 0; i <= s-1; i++){ if

犯了一个很脑残的错误，忘记return 0 了，总是运行错误#includeint main(void){ int n, i, a; scanf("%d", &n); int min = 10000, max = -10000, sum = 0; for( i = 1; i   scanf("%d", &a);  if( a > max) max = a;  i

#includeint main(void){ long long n; scanf("%X", &n); printf("%u\n", n); return 0;}呵呵O(∩_∩)O以下是不那么猥琐的常规解法。。。一道这么简单的题递交八次也是醉醉的#include#include#includeint main(void){ long long n = 0; c

#includeint main(void){ int a, n, m, o; for( a = 123; a <= 999; a++){ n = a / 100; m = a % 100 /10; o = a % 10; if( a == n*n*n + m*m*m + o*o*o){ printf("%d\n", a); } } return 0;}

#include#include#includeint main(void){ char a[11][5]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu","shi"}; char b[10]; char c[5][5] = {"\n","shi", "bai","qian"}; char d[3][

我觉得我的太过麻烦#include#include#includeint main(void){ har a[11], b[11]; int s1, s2, i, flag; gets(a); gets(b); s1 = strlen(a); s2 = strlen(b); if( s1 != s2){ printf("1"); } if( s1 == s2){

#includeint main(void){ int n, a, b, c; scanf("%d", &n); for(a = 1; a <= 9; a++){ for( b = 0; b <= 9; b++){ c = n - (2 * a + 2 * b); if( c < 0){ break; } else if(c < 10){ pri

#include#include#includeint main(void){ int m, i, j, temp; int a[205]; scanf("%d", &m); for( i = 1; i <= m; i++){ scanf("%d", &a[i]); } // 冒泡排序法 for( i = m; i >= 1; i--){ for( j = 1; j <

蓝桥杯题库的一些做题答案，主要是留下一个记录，以便日后的提高