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摘要： 简单题View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;#define maxl 1005#define maxn 30int f[maxn], g[maxn];char a[maxl], b[maxl];void make(char *st, int f[]){ int len = strlen(st); for (int i = 0; i < len; i++) f[st[i] - & 阅读全文

摘要： 简单题View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;struct Point{ double x, y;} point[5];void input(){ for (int i = 1; i < 4; i++) scanf("%lf%lf", &point[i].x, &point[i].y); if (point[1].x != point[2].x 阅读全文

摘要： 题意：给出p,q,r,s，求c(p,q)/c(r,s)分析：利用组合数公式，写出整体的分子和分母（用数组标记，分子分母各有哪些数），把相同的数字约去。然后把分子乘上，分母除掉。但是这样直接乘除会损失精度。所以我们定义double ans = 1;然后当ans<1时算乘法，ans>=1时算除法，以控制ans始终在1不会和1相差10000倍以上，以保证精度。View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using name 阅读全文

摘要： 最短路View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;#define maxf 105#define maxn 505#define inf 0x3f3f3f3fint firenum, n;int fire[maxf];int map[maxn][maxn];int firedist[maxn];void input(){ scanf("%d%d", &firen 阅读全文

摘要： 打表View Code #include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;#define maxn 35005int n;int pre[maxn], next[maxn];void del(int a){ pre[next[a]] = pre[a]; next[pre[a]] = next[a];}void init(){ n = 35000; for (int i = 0; i < n; i++) { p 阅读全文