Windrunner

输入数据31117532写成31117531，导致我调了2小时，你妹。。思想是化整为零。。。#include#include#include#include#include#include#includeusing namespace std;const int maxN=19;int cat[maxN],sum[maxN+1];void build(int d,

纯数学题，算出空间一点到一空间直线的距离由题目描述可写出直线的点向式方程(x-x2)/a=(y-y2)/b=(z-z2)/c=tt然后算出于与该直线垂直的该点所在平面，(a,b,c)为该平面的法向量，因此可得a*(x-x1)+b*(y-y1)+c*(z-z1)=0算出空间直线与平面相交的点a*(a*tt+x2-x1)+b*(b*tt+y2-y1)+c*(c*tt+z2-z1)=0该

#include#include#include#include#include#include#include#include#includeusing namespace std;int main(){ char s2[100],s[20][100]; int i,j,cnt[100],x,y,loop; bool flag1; while(scanf("%

你妹的模拟题。。。wa了n次#include#include#include#include#include#include#includeusing namespace std;int main(){#ifndef ONLINE_JUDGE freopen("a.in","r",stdin);#endif int a1,a2,a3,a4,a5,a6;

设答案为x，y则距离的运算公式为|x-x1|+|x-x2|+...+|x-xn|+|y-y1|+|y-y2|+...+|y-yn|因此当x，y均为中位数时上式最小用cin耗时2687ms用scanf耗时453ms#include#include#include#include#include#includeusing namespace std;const i

#include#include#includechar s[20],a[20];bool vis[20];int len,flag;void solve(int d){ int i; if(d!=len) for(i=0;i<len;i++) { if(!vis[i]&&(flag||a[d]!=s[i])) {

#include#include#includeint main(){ int i,n,N,a,b,x[110],y[110],cnt; while(scanf("%d",&N)&&N) { n=2*N; for(i=0;i<n;i++) scanf("%d%d",&x[i],&y[i]); f

我会告诉你我把这道题当模拟题写的么。。。坑爹啊#include#include#include#include#include#include#include#include#includeusing namespace std;int a[100000];int main(){ int n,m,i,ans; double t=sqrt(10000000

二分答案#include #include #include #include #include #include using namespace std;int a[10010];int main(){#ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w