Codeforces Round #427 (Div. 2) （A B C）

A. Key races

Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of s characters. The first participant types one character in v₁ milliseconds and has ping t₁ milliseconds. The second participant types one character in v₂ milliseconds and has ping t₂ milliseconds.

If connection ping (delay) is t milliseconds, the competition passes for a participant as follows:

1. Exactly after t milliseconds after the start of the competition the participant receives the text to be entered.
2. Right after that he starts to type it.
3. Exactly t milliseconds after he ends typing all the text, the site receives information about it.

The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.

Given the length of the text and the information about participants, determine the result of the game.

Input

The first line contains five integers s, v₁, v₂, t₁, t₂ (1 ≤ s, v₁, v₂, t₁, t₂ ≤ 1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.

Output

If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".

Examples

Input

5 1 2 1 2

Output

First

Input

3 3 1 1 1

Output

Second

Input

4 5 3 1 5

Output

Friendship

看谁用时少就行。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main() {
 5     int s, v1, v2, t1, t2;
 6     cin>>s>>v1>>v2>>t1>>t2;
 7     int a = v1*s+2*t1, b = v2*s+2*t2;
 8     if(a < b)printf("First\n");
 9     else if(a == b) printf("Friendship\n");
10     else printf("Second\n");
11     return 0;
12 }

B. The number on the board

Some natural number was written on the board. Its sum of digits was not less than k. But you were distracted a bit, and someone changed this number to n, replacing some digits with others. It's known that the length of the number didn't change.

You have to find the minimum number of digits in which these two numbers can differ.

Input

The first line contains integer k (1 ≤ k ≤ 10⁹).

The second line contains integer n (1 ≤ n < 10¹⁰⁰⁰⁰⁰).

There are no leading zeros in n. It's guaranteed that this situation is possible.

Output

Print the minimum number of digits in which the initial number and n can differ.

Examples

Input

3
11

Output

1

Input

3
99

Output

0

Note

In the first example, the initial number could be 12.

In the second example the sum of the digits of n is not less than k. The initial number could be equal to n.

 

 

从小到大排下序，每次改变最小的数就行。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int MAX = 100010;
 4 char str[MAX];
 5 int main() {
 6     int k;
 7     cin >> k >> str;
 8     int sum = 0, len = strlen(str);
 9     for(int i = 0; i < len; i ++) {
10         sum += str[i] - '0';
11     }
12     int ans = 0;
13     sort(str,str+len);
14     if(sum >= k) return 0*printf("0\n");
15     for(int i = 0; i < len; i ++) {
16         if(str[i] != '9') {
17             sum += '9' - str[i];
18             ans++;
19             if(sum >= k)return 0*printf("%d\n",ans);
20         }
21     }
22     return 0;
23 }

C. Star sky

The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (x_i, y_i), a maximum brightness c, equal for all stars, and an initial brightness s_i (0 ≤ s_i ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness s_i. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment t_i and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x_1i, y_1i) and the upper right — (x_2i, y_2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 10⁵, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers x_i, y_i, s_i (1 ≤ x_i, y_i ≤ 100, 0 ≤ s_i ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers t_i, x_1i, y_1i, x_2i, y_2i (0 ≤ t_i ≤ 10⁹, 1 ≤ x_1i < x_2i ≤ 100, 1 ≤ y_1i < y_2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

Examples

Input

2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5

Output

3
0
3

Input

3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51

Output

3
3
5
0

注意，在一个坐标上可以有两个或两个以上亮度相同的点。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int a[110][110][110];
 4 int main() {
 5     int n,q,c;
 6     cin>>n>>q>>c;
 7     for(int i = 1; i <= n; i ++) {
 8         int x, y, s;
 9         cin>>x>>y>>s;
10         for(int j = 0; j <= c; j ++) {
11             a[j][x][y] += (s+j)%(c+1);
12         }
13     }
14     for(int i = 1; i <= 100; i ++) {
15         for(int j = 1; j <= 100; j ++) {
16             for(int k = 0; k <= c; k ++) {
17                 a[k][i][j] += a[k][i-1][j]+a[k][i][j-1]-a[k][i-1][j-1];
18             }
19         }
20     }
21     while(q--) {
22         int x1,x2,y1,y2,t;
23         cin>>t>>x1>>y1>>x2>>y2;
24         t%=(c+1);
25         printf("%d\n",a[t][x2][y2]-a[t][x2][y1-1]-a[t][x1-1][y2]+a[t][x1-1][y1-1]);
26     }
27     return 0;
28 }

 

转载于:https://www.cnblogs.com/xingkongyihao/p/7267452.html