2015
04-14

# Harmonious Set

For a giving integer n ( n > 0 ) , the set Sn consists of the non negative integers less than n. For example:S5 = {0,1,2,3,4}. A subset of Sn is harmonious if and only if the sum of its elements is a multiply of n. Now your task is easy. For a given n , you should find the number of harmonious subset of Sn.

There is a number C in the first line , meaning there are C cases . C is guaranteed no more than 300.

Then C cases below. Each case is a positive integer n in a single line. n is not greater than 10^9.

There is a number C in the first line , meaning there are C cases . C is guaranteed no more than 300.

Then C cases below. Each case is a positive integer n in a single line. n is not greater than 10^9.

5
1
2
3
10
1000

2
2
4
104
618918635

/*代码慢慢敲，教训*/

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define size 1010
#define INF 99999999
int n,m,a,b,d,p,s,t;
int S[size],dist[size],value[size];
int map[size][size],cos[size][size];
void dijkstra(int u0)
{

for(int i=1;i<=n;i++)
{
dist[i]=map[u0][i];
value[i]=cos[u0][i];
S[i]=0;
}
S[u0]=1;
for(int i=1;i<n;i++)
{
int min=INF,v=u0;
for(int j=1;j<=n;j++)
{
if(min>dist[j] && !S[j])
{
v=j,min=dist[j];
}
}
S[v]=1;
for(int j=1;j<=n;j++)
{
if(!S[j] && map[v][j]<INF)
{
if(dist[j]>dist[v]+map[v][j])
{
dist[j]=dist[v]+map[v][j];
value[j]=value[v]+cos[v][j];
}
else if(dist[j]==dist[v]+map[v][j])
{
if(value[j]>value[v]+cos[v][j])
{
value[j]=value[v]+cos[v][j];
}
}
}

}
}
printf("%d %d\n",dist[t],value[t]);
}
int main()
{
while(~scanf("%d%d",&n,&m)&&n+m)
{
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
map[i][j]=INF;
cos[i][j]=INF;
}
for(int i=0;i<m;i++)
{
scanf("%d%d%d%d",&a,&b,&d,&p);
{
if(map[a][b]>d)
{
map[a][b]=map[b][a]=d;
cos[a][b]=cos[b][a]=p;
}
else if(map[a][b]==d)//该死的等于，代码要慢慢敲，经验
{
if(cos[a][b]>p)
{
cos[a][b]=cos[b][a]=p;
}
}
}
}
scanf("%d%d",&s,&t);
dijkstra(s);
}
return 0;
}

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