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A. Alyona and copybooks
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputLittle girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks. What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase. Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109). Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4. Examples
input
1 1 3 4 output
3 input
6 2 1 1 output
1 input
4 4 4 4 output
0 input
999999999 1000000000 1000000000 1000000000 output
1000000000 Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally. In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total. In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything. In the fourth example Alyona should buy one pack of one copybook. 题意:1件的费用为a 2件的费用为b 3件的费用为c 现在已经拥有n件物品 购买另外k件 使得(n+k)%4==0 输出最小的花费 题解;水 共有9种策略 1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #define ll __int64 5 #define mod 10000000007 6 using namespace std; 7 ll n,a,b,c; 8 int main() 9 { 10 scanf("%I64d %I64d %I64d %I64d",&n,&a,&b,&c); 11 if(n%4==0) 12 { 13 printf("0\n"); 14 } 15 else 16 { 17 int exm=4-n%4; 18 if(exm==1) 19 { 20 printf("%I64d\n",min(a,min(b+c,3*c))); 21 } 22 if(exm==2) 23 { 24 printf("%I64d\n",min(2*a,min(b,c*2))); 25 } 26 if(exm==3) 27 { 28 printf("%I64d\n",min(3*a,min(b+a,c))); 29 } 30 } 31 return 0; 32 }
B. Alyona and flowers
time limit per test
2 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputLittle Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative. Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in. For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this! Alyona can choose any number of the subarrays, even 0 or all suggested by her mother. Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother. The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100). The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri(1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri]. Each subarray can encounter more than once. Output
Print single integer — the maximum possible value added to the Alyona's happiness. Examples
input
5 4 output
7 input
4 3 output
16 input
2 2 output
0 Note
The first example is the situation described in the statements. In the second example Alyona should choose all subarrays. The third example has answer 0 because Alyona can choose none of the subarrays. 题意:给你n个数 m个区间 在这m个区间中取若干个区间 对于选择的区间求和累加 输出最大值 题解:水 对于某个区间 若区间和大于0则累加否则舍弃 1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #define ll __int64 6 #define mod 10000000007 7 using namespace std; 8 int n,m; 9 int a[105]; 10 int sum[105]; 11 int b[105]; 12 int l,r; 13 int main() 14 { 15 scanf("%d %d",&n,&m); 16 sum[0]=0; 17 for(int i=1;i<=n;i++) 18 { 19 scanf("%d",&a[i]); 20 sum[i]=sum[i-1]+a[i]; 21 } 22 for(int i=1;i<=m;i++) 23 { 24 scanf("%d %d",&l,&r); 25 b[i]=sum[r]-sum[l-1]; 26 } 27 sort(b+1,b+1+m); 28 int ans=0; 29 for(int i=m;i>=1;i--) 30 ans=ans+max(0,b[i]); 31 printf("%d\n",ans); 32 return 0; 33 }
C. Alyona and mex
time limit per test
2 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputAlyona's mother wants to present an array of n non-negative integers to Alyona. The array should be special. Alyona is a capricious girl so after she gets the array, she inspects m of its subarrays. Subarray is a set of some subsequent elements of the array. The i-th subarray is described with two integers li and ri, and its elements are a[li], a[li + 1], ..., a[ri]. Alyona is going to find mex for each of the chosen subarrays. Among these m mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible. You are to find an array a of n elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible. The mex of a set S is a minimum possible non-negative integer that is not in S. Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105). The next m lines contain information about the subarrays chosen by Alyona. The i-th of these lines contains two integers li and ri(1 ≤ li ≤ ri ≤ n), that describe the subarray a[li], a[li + 1], ..., a[ri]. Output
In the first line print single integer — the maximum possible minimum mex. In the second line print n integers — the array a. All the elements in a should be between 0 and 109. It is guaranteed that there is an optimal answer in which all the elements in a are between 0 and 109. If there are multiple solutions, print any of them. Examples
input
5 3 output
2 input
4 2 output
3 Note
The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5)is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2. 题意:要求你构造一个长度为n的数列 给你m个区间 某个区间的mex值=当前区间内没有出现过的最小的非负数 要求构造的数列中 区间mex的最小值尽可能大 题解:找到给定区间的最小长度 len 构造一个 0 1 2...len-1 0 1 2...len-1...的数列 满足题目要求 并且mex值的最小值为len 为说明问题 对于题目的样例一 5 3 1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #define ll __int64 6 #define mod 10000000007 7 using namespace std; 8 int n,m; 9 int ans[100005]; 10 int l[100005],r[100005]; 11 int main() 12 { 13 scanf("%d %d",&n,&m); 14 int exm=200000; 15 for(int i=1;i<=m;i++) 16 { 17 scanf("%d %d",&l[i],&r[i]); 18 exm=min(exm,r[i]-l[i]); 19 } 20 for(int i=1;i<=n;i++) 21 ans[i]=0; 22 for(int i=1;i<=n;i++) 23 { 24 for(int j=0;j<=exm;j++) 25 { 26 if(i+j<=n) 27 { 28 ans[i+j]=j; 29 } 30 } 31 i=i+exm; 32 } 33 printf("%d\n",exm+1); 34 for(int i=1;i<=n;i++) 35 printf("%d ",ans[i]); 36 printf("\n"); 37 return 0; 38 }
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