CodeForces 300C 最短路

A

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 300C

Description

Vitaly is a very weird man. He's got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal representation of this integer only contains digits a and b. Vitaly calls a good number excellent, if the sum of its digits is a good number.

For example, let's say that Vitaly's favourite digits are 1 and 3, then number 12 isn't good and numbers 13 or 311 are. Also, number111 is excellent and number 11 isn't.

Now Vitaly is wondering, how many excellent numbers of length exactly n are there. As this number can be rather large, he asks you to count the remainder after dividing it by 1000000007(10⁹ + 7).

A number's length is the number of digits in its decimal representation without leading zeroes.

Input

The first line contains three integers: a, b, n(1 ≤ a < b ≤ 9, 1 ≤ n ≤ 10⁶).

Output

Print a single integer — the answer to the problem modulo 1000000007(10⁹ + 7).

Sample Input

Input

1 3 3

Output

1

Input

2 3 10

Output

165

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
struct Node
{
    int x,y;
} sta[100005];
int map[105][105];
int main()
{
    int n,d;
    int tim[100005];
    scanf("%d%d",&n,&d);
    for(int i=2; i<=n-1; i++)
        scanf("%d",&tim[i]);
    for(int i=1; i<n+1; i++)
        scanf("%d%d",&sta[i].x,&sta[i].y);
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            if(i!=j)map[i][j]=99999999;
            else map[i][j]=0;
    //cout<<map[2][3];
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
        {
            if(i!=j)
            {
                map[i][j]=(abs(sta[i].x-sta[j].x)+abs(sta[i].y-sta[j].y))*d-tim[j];

            }
        }
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                map[i][j]=min(map[i][j],map[i][k]+map[k][j]);
            }
    //for(int i=1;i<=n;i++)
      printf("%d\n",map[1][n]);
}
/*
6 1000
142 712 254 869
7 0
95 38
96 -20
-7 93
75 -45
-80 -20
*/