A. Lucky Year
time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output
Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.
You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.
Input
The first line contains integer number n (1 ≤ n ≤ 109) — current year in Berland.
Output
Output amount of years from the current year to the next lucky one.
Examples
Input
代码语言：javascript复制4Output
代码语言：javascript复制1Input
代码语言：javascript复制201Output
代码语言：javascript复制99Input
代码语言：javascript复制4000Output
代码语言：javascript复制1000Note
In the first example next lucky year is 5. In the second one — 300. In the third — 5000.
题目链接：http://codeforces.com/contest/808/problem/A
分析：题目意思是要将下一个数字变为只有1个非0的数，打表即可，要注意开ll，否则会WA！
下面给出AC代码：
代码语言：javascript复制 1 #include <bits/stdc .h>
2 using namespace std;
3 typedef long long ll;
4 int main()
5 {
6 ll n;
7 while(scanf("%lld",&n)!=EOF)
8 {
9 if(n>=0&&n<=9)
10 cout<<1<<endl;
11 else if(n>=10&&n<=99)
12 cout<<(n/10 1)*10-n<<endl;
13 else if(n>=100&&n<=999)
14 cout<<(n/100 1)*100-n<<endl;
15 else if(n>=1000&&n<=9999)
16 cout<<(n/1000 1)*1000-n<<endl;
17 else if(n>=10000&&n<=99999)
18 cout<<(n/10000 1)*10000-n<<endl;
19 else if(n>=100000&&n<=999999)
20 cout<<(n/100000 1)*100000-n<<endl;
21 else if(n>=1000000&&n<=9999999)
22 cout<<(n/1000000 1)*1000000-n<<endl;
23 else if(n>=10000000&&n<=99999999)
24 cout<<(n/10000000 1)*10000000-n<<endl;
25 else if(n>=100000000&&n<=999999999)
26 cout<<(n/100000000 1)*100000000-n<<endl;
27 else if(n>=1000000000&&n<=9999999999)
28 cout<<(n/1000000000 1)*1000000000-n<<endl;
29 }
30 return 0;
31 }B. Average Sleep Time
time limit per test：1 second
memory limit per test：256 megabytes
input：standard input
output：standard output
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts k days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last n days. So now he has a sequence a1, a2, ..., an, where ai is the sleep time on the i-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider k consecutive days as a week. So there will be n - k 1 weeks to take into consideration. For example, if k = 2, n = 3 and a = [3, 4, 7], then the result is
.
You should write a program which will calculate average sleep times of Polycarp over all weeks.
Input
The first line contains two integer numbers n and k (1 ≤ k ≤ n ≤ 2·105).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).
Output
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10 - 6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
Examples
Input
代码语言：javascript复制3 2
3 4 7Output
代码语言：javascript复制9.0000000000Input
代码语言：javascript复制1 1
10Output
代码语言：javascript复制10.0000000000Input
代码语言：javascript复制8 2
1 2 4 100000 123 456 789 1Output
代码语言：javascript复制28964.2857142857Note
In the third example there are n - k 1 = 7 weeks, so the answer is sums of all weeks divided by 7.
题目链接：http://codeforces.com/contest/808/problem/B
分析：计算上升和下降的序列数之和，再除以n-k 1即可，相当于求其中每k个数的和的平均值！
下面给出AC代码：
代码语言：javascript复制 1 #include <bits/stdc .h>
2 typedef long long ll;
3 using namespace std;
4 ll n,k;
5 double avg;
6 ll ans;
7 ll s[200005];
8 int main()
9 {
10 cin>>n>>k;
11 for(int i=1;i<=n;i )
12 scanf("%I64d",s i);
13 ll c=0;
14 for(int i=1;i<=n;i )
15 {
16 if(i<=n-k 1)
17 {
18 if(c<k)
19 c ;
20 ans =c*s[i];
21 }
22 else
23 {
24 if(c>n-i 1)
25 c--;
26 ans =c*s[i];
27 }
28 }
29 avg=ans*1.0/(n-k 1);
30 printf("%0.12lf",avg);
31 return 0;
32 }C. Tea Party
time limit per test：1 second
memory limit per test：256 megabytes
input：standard input
output：standard output
Polycarp invited all his friends to the tea party to celebrate the holiday. He has n cups, one for each of his n friends, with volumes a1, a2, ..., an. His teapot stores w milliliters of tea (w ≤ a1 a2 ... an). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume
- Every cup will contain integer number of milliliters of tea
- All the tea from the teapot will be poured into cups
- All friends will be satisfied.
Friend with cup i won't be satisfied, if there exists such cup j that cup i contains less tea than cup j but ai > aj.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input
The first line contains two integer numbers n and w (1 ≤ n ≤ 100,
).
The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 100).
Output
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Examples
Input
代码语言：javascript复制2 10
8 7Output
代码语言：javascript复制6 4 Input
代码语言：javascript复制4 4
1 1 1 1Output
代码语言：javascript复制1 1 1 1 Input
代码语言：javascript复制3 10
9 8 10Output
代码语言：javascript复制-1Note
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
题目链接：http://codeforces.com/contest/808/problem/C
题意：n个茶杯，每个茶杯有容量。现在给一壶茶，总量为w。希望倒茶满足条件：每杯茶要超过容量的一半，并且w被倒光，茶杯内的茶水为整数，容量大的杯子内的茶不允许比容量小的杯子内的茶水少，特判不满足情况，然后将茶水给每一杯倒至一半以上。然后按照容量从大到小每次倒一个单位，直到倒完为止。
分析：贪心求解！ 下面给出AC代码：
代码语言：javascript复制 1 #include <bits/stdc .h>
2 using namespace std;
3 class Tea
4 {
5 public:
6 int a,w,id;
7 };
8 Tea t[105];
9 int n,w;
10 int cmp1(Tea a,Tea b)
11 {
12 return a.a>b.a;
13 }
14 int cmp2(Tea a,Tea b)
15 {
16 return a.id<b.id;
17 }
18 int main()
19 {
20 cin>>n>>w;
21 for(int i=0;i<n;i )
22 {
23 cin>>t[i].a;
24 t[i].id=i;
25 }
26 sort(t,t n,cmp1); //杯子大到小排序
27 for(int i=0;i<n;i )
28 {
29 w-=(t[i].a 1)/2; //每个杯子至少装一半
30 t[i].w=(t[i].a 1)/2; //保存已经装了多少
31 }
32 for(int i=0;i<n;i )
33 {
34 int x=t[i].a-t[i].w;
35 if(w>=x)
36 {
37 t[i].w =x;
38 w-=x;
39 }
40 else if(w>0)
41 {
42 t[i].w =w;
43 w=0;
44 }
45 else
46 break;
47 }
48 sort(t,t n,cmp2); //按照id 排序 排回原位
49 if(w<0)
50 cout<<-1;
51 else
52 for(int i=0;i<n;i )
53 cout<<t[i].w<<" ";
54 return 0;
55 }
