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摘要： 简单题View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;#define maxl 20char st1[maxl], st2[maxl];int getd(char *st, char a){ int len = strlen(st); int x = -1; for (int i = 0; i < len; i++) if (st[i] == a) { x = i; ... 阅读全文

摘要： 简单题View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;int t, x, m;int n;void input(){ scanf("%d%d%d", &t, &x, &m); n = 0x3f3f3f3f; for (int i = 0; i < m; i++) { int a, b; scanf("%d%d", & 阅读全文

摘要： 题意：给出d,n，求d/1+d/2+……+d/n，所有除法向上取整。分析：我们从两边同时计算，算d/i时，同时算出由面所有d/x等于i的项的和，直到两端汇合。这些x中的一个应该是minx = d/i向上取整，因为minx*i>=d且(minx-1)*i<d，所以d/minx<=i且d/(minx-1)>i。也就是说minx再小一点那么d/minx向上取整的结果就会比i大。复杂度为sqrt(n)。View Code #include <iostream>#include <cstdlib>#include <cstring>#inclu 阅读全文

摘要： 简单题View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int n;int f[] ={ 3, 5, 8, 12, 20, 34, 57, 98, 170, 300, 536, 966, 1754, 3210, 5910, 10944, 20366, 38064, 71421, 134480, 254016};int main(){ //freopen 阅读全文

摘要： 题意：这题的本质是给定若干条边，问能否用其中的一些组成多边形。分析：先排序，看前i个的和是否大于第i+1个，若大于则可以View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;#define maxn 30double f[maxn];int n;void input(){ for (int i = 0; i < n; i++) scanf(& 阅读全文