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简单题#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;int a, b, c, d;int cal(int a){ return (a + 40) % 40;}int main(){ while (scanf("%d%d%d%d", &a, &b, &c, &d), a | b | c | d) { int ans = cal(a - b) + cal(c - 阅读全文
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简单题#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>using namespace std;#define maxn 1005int n;int f[maxn];double average;void input(){ int sum = 0; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", &f[i]); sum += 阅读全文
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简单题#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>using namespace std;#define maxn 1000005char st[maxn];int len;bool ok(int a){ for (int i = a; i < len; i++) if (st[i] != st[i % a]) return false; return true;}int work(){ for (int i = 1; i <= len 阅读全文
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几何公式#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <cmath>using namespace std;#define pi acos(-1)int D, V;int main(){ while (scanf("%d%d", &D, &V), D | V) { printf("%.3f\n", pow(D * D * D - 6 * V / pi, 1.0 / 阅读全文
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题意：ｎ<=10000，并保证存在ｎ的某个倍数的十进制表示形式各个数位都为1。问这个数是ｎ的多少倍。分析：我们枚举1,11,111……直到找到能被ｎ整除的为止。为了防止这个1序列过大，我们不断将其对ｎ取余，这样可以保证其一直在int范围内，并且不影响整除性。#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>using namespace std;int n;int main(){ while (scanf("%d", & 阅读全文