B. Marathon
time limit per test：1 second
memory limit per test：256 megabytes
input：standard input
output：standard output
Valera takes part in the Berland Marathon. The marathon race starts at the stadium that can be represented on the plane as a square whose lower left corner is located at point with coordinates (0, 0) and the length of the side equals a meters. The sides of the square are parallel to coordinate axes.
As the length of the marathon race is very long, Valera needs to have extra drink during the race. The coach gives Valera a bottle of drink each d meters of the path. We know that Valera starts at the point with coordinates (0, 0) and runs counter-clockwise. That is, when Valera covers a meters, he reaches the point with coordinates (a, 0). We also know that the length of the marathon race equalsnd + 0.5 meters.
Help Valera's coach determine where he should be located to help Valera. Specifically, determine the coordinates of Valera's positions when he covers d, 2·d, ..., n·d meters.
The first line contains two space-separated real numbers a and d (1 ≤ a, d ≤ 105), given with precision till 4 decimal digits after the decimal point. Number a denotes the length of the square's side that describes the stadium. Number d shows that after each d meters Valera gets an extra drink.
The second line contains integer n (1 ≤ n ≤ 105) showing that Valera needs an extra drink n times.
Print n lines, each line should contain two real numbers xi and yi, separated by a space. Numbers xi and yi in the i-th line mean that Valera is at point with coordinates (xi, yi) after he covers i·d meters. Your solution will be considered correct if the absolute or relative error doesn't exceed 10 - 4.
Note, that this problem have huge amount of output data. Please, do not use cout stream for output in this problem.
2 5
2
1.0000000000 2.0000000000
2.0000000000 0.0000000000
4.147 2.8819
6
2.8819000000 0.0000000000
4.1470000000 1.6168000000
3.7953000000 4.1470000000
0.9134000000 4.1470000000
0.0000000000 2.1785000000
0.7034000000 0.0000000000
这个题太坑爹了。。。。。WA了好几遍，原来就是因为该用long long我用了int。。。。。。这个题好想，把这个正方形扯成直线再去做，因为浮点数不能用普通的取余，所以要用别的办法，转化成别的类型取余，或者是直接减也行，还可以用fmod，膜拜啊，我今天才知道这个函数、
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <math.h> #include <stdlib.h> #include <map> using namespace std ; const int INF = 999999999 ; const double eps = 1e-8 ; double ch[100010] ; int main() { double a,d ; int n ; while(~scanf("%lf %lf",&a,&d)) { scanf("%d",&n) ; double sum = 0.0 ; long long x; for(int i = 1 ; i <= n ; i++) { ch[i] = d*i ; if(ch[i] > 4*a)//也可以把底下这两句变成 ch[i] = fmod(d*i,4*a) ;也对 { x = (long long)(ch[i]/(4*a)) ;//就是这里用了int结果精度废了 ch[i] = ch[i]-x*1.0*4*a ; } } for(int i = 1 ; i <= n ; i++) { if(ch[i] >= 0 && ch[i] <= a) printf("%.9lf %.9lf\n",ch[i],sum) ; else if(ch[i] > a && ch[i] <= 2*a) printf("%.9lf %.9lf\n",a,ch[i]-a) ; else if(ch[i] > 2*a && ch[i] <= 3*a) printf("%.9lf %.9lf\n",3*a-ch[i],a) ; else if(ch[i] > 3*a && ch[i] <= 4*a) printf("%.9lf %.9lf\n",sum,4*a-ch[i]) ; } } return 0 ; }
C. Restore Graph
time limit per test：1 second
memory limit per test：256 megabytes
input：standard input
output：standard output
Valera had an undirected connected graph without self-loops and multiple edges consisting of n vertices. The graph had an interesting property: there were at most k edges adjacent to each of its vertices. For convenience, we will assume that the graph vertices were indexed by integers from 1 to n.
One day Valera counted the shortest distances from one of the graph vertices to all other ones and wrote them out in array d. Thus, element d[i] of the array shows the shortest distance from the vertex Valera chose to vertex number i.
Then something irreparable terrible happened. Valera lost the initial graph. However, he still has the array d. Help him restore the lost graph.
The first line contains two space-separated integers n and k (1 ≤ k < n ≤ 105). Number n shows the number of vertices in the original graph. Number k shows that at most k edges were adjacent to each vertex in the original graph.
The second line contains space-separated integers d[1], d[2], ..., d[n] (0 ≤ d[i] < n). Number d[i] shows the shortest distance from the vertex Valera chose to the vertex number i.
If Valera made a mistake in his notes and the required graph doesn't exist, print in the first line number -1. Otherwise, in the first line print integer m (0 ≤ m ≤ 106) — the number of edges in the found graph.
In each of the next m lines print two space-separated integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting the edge that connects vertices with numbers ai and bi. The graph shouldn't contain self-loops and multiple edges. If there are multiple possible answers, print any of them.
3 2
0 1 1
3
1 2
1 3
3 2
4 2
2 0 1 3
3
1 3
1 4
2 3
3 1
0 0 0
思路 ：这个题比赛的时候没有时间看，后来是问的啸爷，建一棵最小的树，需要注意几个地方就可以了，就是距离为0的点只能有一个，因为只可以有一个根节点，如果距离为 i（ i 不等于1） 的点最多可以有距离为i-1的点乘上k-1个，因为距离为i-1的点还需要有一条边分给父节点，当然，当i为1时，他的点最多可以有m个，因为他的父节点本身就是根节点，不需要再分给父节点的父节点。然后就是存边了
#include <iostream> #include <algorithm> #include <cstdlib> #include <cstdio> #include <cstring> #define _LL __int64 using namespace std; int dis[100100]; _LL ans[100100]; struct N { int v,next; }edge[100100]; int head[100100];//head[i]代表的是所有与选定的点距离为i的点 int Top; int de[100100]; void Link(int u,int v) { edge[Top].v = v;//v存的是该点的下标 edge[Top].next = head[u]; head[u] = Top++; } void Output(int len,int m) { int p ; for(p = head[len];p != -1;p = edge[p].next) { printf("%d %d\n",edge[head[len-1]].v,edge[p].v);//如果距离为len的点应该与距离为len-1的点相连，这样他自己的距离才是（len-1）+1 de[edge[head[len-1]].v]++; if(de[edge[head[len-1]].v] == m)//如果距离为len的这个点的父节点的度数已经到达了，就要换另外一个距离为len-1的点 head[len-1] = edge[head[len-1]].next; de[edge[p].v] = 1;//因为p点已经与其父节点连接了，所以度数要变为1而不是0. } } int main() { int n,i; _LL m; scanf("%d %I64d",&n,&m); memset(head,-1,sizeof(head)); memset(ans,0,sizeof(ans)); memset(de,0,sizeof(de)); Top = 0; int Max = -1; for(i = 1;i <= n; ++i) { scanf("%d",&dis[i]); Max = max(Max,dis[i]); ans[dis[i]]++; Link(dis[i],i); } if(ans[0] != 1) { printf("-1\n"); return 0; } for(i = 1;i <= Max; ++i) { if(ans[i-1] == 0 || (i != 1 && ans[i] > ans[i-1]*(m-1)) || (i == 1 && ans[i] > m) ) { printf("-1\n"); return 0; } } printf("%d\n",n-1); for(i = 1;i <= Max; ++i) { Output(i,m); } return 0; }
#include <stdio.h> #include <iostream> #include <vector> using namespace std ; vector<int>dis[100010] ; vector<pair<int,int> >map ; int main() { int n, k ; while(~scanf("%d %d",&n,&k)) { for(int i = 0 ; i <= n ; i++) dis[i].clear() ; int x ,maxx = 0; for(int i = 1 ; i <= n ; i++) { scanf("%d",&x) ; dis[x].push_back(i) ;//将这个距离和下标存储下来 maxx = max(maxx,x) ;//求最长距离 } if(dis[0].size() != 1)//距离为0的点只能有一个就是它本身，只有一个才能做为根节点，否则多个根节点就不符合 { printf("-1\n"); return 0 ; } for(int i = 1 ; i <= maxx ; i++) { int edge = (i != 1) ,cnt = 0; //当距离为1的时候，不用考虑父节点与其父节点的那条边，因为本身就是根节点，没有父节点，edge为0 //cnt表示的是距选定点距离为i-1的点中的第cnt个 for(int j = 0 ; j < dis[i].size() ; j++)//所有与题目中某点距离为i的点 { if(edge == k)//如果这个点的度数已经到达给出的最大度数k { edge = (i != 1) ; cnt++ ;//就换到下一个点 } if(cnt == dis[i-1].size())//如果所有距离为i的点都用完了 { printf("-1\n") ; return 0 ; } map.push_back(make_pair(dis[i-1][cnt],dis[i][j])) ;//距离为i的点中的第j个与距离为i-1的点中第cnt个相连 edge++ ; } } printf("%d\n",n-1) ;//如果构造的是一棵最小的树，所以n个点最少有n-1条边 for(int i = 0 ; i < map.size() ; i++) printf("%d %d\n",map[i].first,map[i].second) ; } return 0 ; }