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:
- Exactly after t milliseconds after the start of the competition the participant receives the text to be entered.
- Right after that he starts to type it.
- 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.
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.
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
5 1 2 1 2
First
3 3 1 1 1
Second
4 5 3 1 5
Friendship
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw.
水题
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <cstdlib> #include <iomanip> #include <cmath> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x<y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define ios() ios::sync_with_stdio(false) #define INF 0x3f3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; int a,b,a1,b1,s; int main() { cin>>s>>a>>b>>a1>>b1; if(a*s+a1*2<b*s+b1*2) puts("First"); else if(a*s+a1*2>b*s+b1*2) puts("Second"); else puts("Friendship"); return 0; }
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.
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.
Print the minimum number of digits in which the initial number and n can differ.
3
11
1
3
99
0
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.
保证所有位置上的数之和大于等于K，找差值排序
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <cstdlib> #include <iomanip> #include <cmath> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x<y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define ios() ios::sync_with_stdio(false) #define INF 0x3f3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; char a[100009]; int b[1000009],k; bool cmp(int x,int y) { return x>y; } int main() { cin>>k; cin>>a; int ans=0; int len=strlen(a); for(int i=0;a[i]!=' ';i++) { ans+=a[i]-'0'; b[i]=9-(a[i]-'0'); } sort(b,b+len,cmp); if(ans>=k) printf("0 "); else { for(int i=0;i<len;i++) { ans+=b[i]; if(ans>=k){printf("%d ",i+1);return 0;} } } return 0; }
The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).
Over time the stars twinkle. At moment 0 the i-th star has brightness si. 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 ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). 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.
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 xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ 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 ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.
For each view print the total brightness of the viewed stars.
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5
3
0
3
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
3
3
5
0
Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
给定n个星星的坐标以及初始亮度，亮度随之时间的增加而增加，但最大值为C，一旦大于C归零从新开始即周期性变化，然后给定一个时间t，和一个矩阵的左上角的坐标和右下角的坐标，问在该时刻矩阵内所有星星的亮度和，最大子矩阵的变形问题
//最大子矩阵和变形 #include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <cstdlib> #include <iomanip> #include <cmath> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x<y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define ios() ios::sync_with_stdio(false) #define INF 0x3f3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; int dp[105][105][105]; int n,q,c,t,x,y,xl,yl,k; void get_matrix() { for(int i=1;i<=100;i++) { for(int j=1;j<=100;j++) { for(int k=0;k<=10;k++) { dp[i][j][k]+=dp[i-1][j][k]+dp[i][j-1][k]-dp[i-1][j-1][k]; } } } } int get_num(int x,int y,int xx,int yy,int k) { return dp[x][y][k]+dp[xx-1][yy-1][k]-dp[xx-1][y][k]-dp[x][yy-1][k]; } int main() { scanf("%d%d%d",&n,&q,&c); memset(dp,0,sizeof(dp)); for(int i=0;i<n;i++) { scanf("%d%d%d",&x,&y,&k); dp[x][y][k]++; } get_matrix(); while(q--) { int ans=0; scanf("%d%d%d%d%d",&t,&x,&y,&xl,&yl); for(int i=0;i<=10;i++) { int val=(t+i)%(c+1); ans+=val*get_num(xl,yl,x,y,i); } printf("%d ",ans); } return 0; }