标签：codeforces div2
链接：http://codeforces.com/contest/479
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on the blackboard. The task was to insert signs of operations ‘+‘ and ‘*‘, and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let‘s consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
Note that you can insert operation signs only between a and b, and between b and c, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It‘s easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print the maximum value that you can get.
The input contains three integers a, b and c, each on a single line (1?≤?a,?b,?c?≤?10).
Print the maximum value of the expression that you can obtain.
1
2
3
9
2
10
3
60
<span style="font-size:14px;">#include <cstdio>
#include <algorithm>
using namespace std;
int max6(int a, int b, int c, int d, int e, int f)
{
return max(max(max(a,b), max(c,d)),max(e,f));
}
int main()
{
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
int a1 = a + b + c;
int a2 = a * b + c;
int a3 = a * (b + c);
int a4 = a * b * c;
int a5 = a + (b * c);
int a6 = (a + b) * c;
int ans = max6(a1, a2, a3, a4, a5, a6);
printf("%d\n", ans);
}
</span>
As you know, all the kids in Berland love playing with cubes. Little Petya has n towers consisting of cubes of the same size. Tower with number i consists of ai cubes stacked one on top of the other. Petya defines the instability of a set of towers as a value equal to the difference between the heights of the highest and the lowest of the towers. For example, if Petya built five cube towers with heights (8, 3, 2, 6, 3), the instability of this set is equal to 6 (the highest tower has height 8, the lowest one has height 2).
The boy wants the instability of his set of towers to be as low as possible. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. Please note that Petya would never put the cube on the same tower from which it was removed because he thinks it‘s a waste of time.
Before going to school, the boy will have time to perform no more than k such operations. Petya does not want to be late for class, so you have to help him accomplish this task.
The first line contains two space-separated positive integers n and k (1?≤?n?≤?100, 1?≤?k?≤?1000) — the number of towers in the given set and the maximum number of operations Petya can perform. The second line contains n space-separated positive integers ai (1?≤?ai?≤?104) — the towers‘ initial heights.
In the first line print two space-separated non-negative integers s and m (m?≤?k). The first number is the value of the minimum possible instability that can be obtained after performing at most k operations, the second number is the number of operations needed for that.
In the next m lines print the description of each operation as two positive integers i and j, each of them lies within limits from 1 to n. They represent that Petya took the top cube from the i-th tower and put in on the j-th one (i?≠?j). Note that in the process of performing operations the heights of some towers can become equal to zero.
If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them.
3 2
5 8 5
0 2
2 1
2 3
3 4
2 2 4
1 1
3 2
5 3
8 3 2 6 3
3 3
1 3
1 2
1 3
In the first sample you need to move the cubes two times, from the second tower to the third one and from the second one to the first one. Then the heights of the towers are all the same and equal to 6.
分析：分能否整除两种情况，每操作一次就排一次序，因为n不大，不会超时，要特判n=1和a[i]都相等的情况
<span style="font-size:14px;">#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 1005;
struct Info
{
int h;
int pos;
}info[MAX];
int re[5000];
int cmp(Info a, Info b)
{
return a.h < b.h;
}
int main()
{
memset(re, 0, sizeof(re));
int n, k, sum = 0, ave = 0, mod = 0;
scanf("%d %d", &n, &k);
for(int i = 1; i <= n; i++)
{
info[i].pos = i;
scanf("%d", &info[i].h);
sum += info[i].h;
}
sort(info + 1, info + n + 1, cmp);
mod = sum % n;
ave = sum / n;
int tmp = 0;
int cnt = 0;
int cnt2 = 0;
int ma = info[n].h - info[1].h;
if(n == 1 || ma == 0)
{
printf("0 0\n");
return 0;
}
while(1)
{
if(mod != 0)
{
k--;
cnt2++;
info[n].h--;
info[1].h++;
re[cnt++] = info[n].pos;
re[cnt++] = info[1].pos;
sort(info + 1, info + n + 1, cmp);
ma = min(ma, info[n].h - info[1].h);
if(ma == 1 || k == 0)
break;
}
else
{
k--;
cnt2++;
info[n].h--;
info[1].h++;
re[cnt++] = info[n].pos;
re[cnt++] = info[1].pos;
sort(info + 1, info + n + 1, cmp);
ma = min(ma, info[n].h - info[1].h);
if(ma == 0 || k == 0)
break;
}
}
printf("%d %d\n", ma, cnt2);
for(int i = 0; i < cnt - 1; i += 2)
printf("%d %d\n", re[i], re[i+1]);
}
</span>
Student Valera is an undergraduate student at the University. His end of term exams are approaching and he is to pass exactly n exams. Valera is a smart guy, so he will be able to pass any exam he takes on his first try. Besides, he can take several exams on one day, and in any order.
According to the schedule, a student can take the exam for the i-th subject on the day number ai. However, Valera has made an arrangement with each teacher and the teacher of the i-th subject allowed him to take an exam before the schedule time on day bi (bi?<?ai). Thus, Valera can take an exam for the i-th subject either on day ai, or on day bi. All the teachers put the record of the exam in the student‘s record book on the day of the actual exam and write down the date of the mark as number ai.
Valera believes that it would be rather strange if the entries in the record book did not go in the order of non-decreasing date. Therefore Valera asks you to help him. Find the minimum possible value of the day when Valera can take the final exam if he takes exams so that all the records in his record book go in the order of non-decreasing date.
The first line contains a single positive integer n (1?≤?n?≤?5000) — the number of exams Valera will take.
Each of the next n lines contains two positive space-separated integers ai and bi (1?≤?bi?<?ai?≤?109) — the date of the exam in the schedule and the early date of passing the i-th exam, correspondingly.
Print a single integer — the minimum possible number of the day when Valera can take the last exam if he takes all the exams so that all the records in his record book go in the order of non-decreasing date.
3
5 2
3 1
4 2
2
3
6 1
5 2
4 3
6
In the first sample Valera first takes an exam in the second subject on the first day (the teacher writes down the schedule date that is 3). On the next day he takes an exam in the third subject (the teacher writes down the schedule date, 4), then he takes an exam in the first subject (the teacher writes down the mark with date 5). Thus, Valera takes the last exam on the second day and the dates will go in the non-decreasing order: 3, 4, 5.
In the second sample Valera first takes an exam in the third subject on the fourth day. Then he takes an exam in the second subject on the fifth day. After that on the sixth day Valera takes an exam in the first subject.
分析：先对给定日期排序，相同的话对提前日期排队，优先考虑提前日期
<span style="font-size:14px;">#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 5005;
struct Info
{
int a, b;
}info[MAX];
int cmp(Info x, Info y)
{
if(x.a == y.a)
return x.b < y.b;
return x.a < y.a;
}
int main()
{
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d %d", &info[i].a, &info[i].b);
sort(info, info + n, cmp);
int ans = info[0].b;
for(int i = 1; i < n; i++)
{
if(ans <= info[i].b)
ans = info[i].b;
else
ans = info[i].a;
}
printf("%d\n", ans);
}</span>
Codeforces Round #274 (Div. 2)
标签：codeforces div2
原文地址：http://blog.csdn.net/tc_to_top/article/details/40278551
