标签：ssis pie ever cer details coff sdi 字符串 erro
A - A 这题很巧妙啊，前两天刚好做过，而且就在博客里
Little C loves number «3» very much. He loves all things about it.
Now he has a positive integer nn. He wants to split nn into 33 positive integers a,b,ca,b,c, such that a+b+c=na+b+c=n and none of the 33 integers is a multiple of 33. Help him to find a solution.
A single line containing one integer nn (3≤n≤1093≤n≤109) — the integer Little C has.
Print 33 positive integers a,b,ca,b,c in a single line, such that a+b+c=na+b+c=n and none of them is a multiple of 33.
It can be proved that there is at least one solution. If there are multiple solutions, print any of them.
3
1 1 1
233
77 77 79
被样例困惑了很久，看了题解啧啧，其实这题就是构造
大致题意：给你一个数n,找到三个数a,b,c，使得a+b+c=n,并且a,b,c都不能是3的倍数
分析：当n%3=0时，n-2一定不是3的倍数，可以构造为a=1,b=1,c=n-2;当n%3!=0时，n-3一定不是3的倍数，那么可以构造为，a=1,b=2,c=n-3.注意这个构造是任意的，只要满足条件即可
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 int main()
6 {
7 int n;
8 while(~scanf("%d",&n))
9 {
10 if(n%3==0)
11 printf("1 1 %d\n",n-2);
12 else
13 printf("1 2 %d\n",n-3);
14 }
15 return 0;
16 }
There are nn points on the plane, (x1,y1),(x2,y2),…,(xn,yn)(x1,y1),(x2,y2),…,(xn,yn).
You need to place an isosceles triangle with two sides on the coordinate axis to cover all points (a point is covered if it lies inside the triangle or on the side of the triangle). Calculate the minimum length of the shorter side of the triangle.
First line contains one integer nn (1≤n≤1051≤n≤105).
Each of the next nn lines contains two integers xixi and yiyi (1≤xi,yi≤1091≤xi,yi≤109).
Print the minimum length of the shorter side of the triangle. It can be proved that it‘s always an integer.
3
1 1
1 2
2 1
3
4
1 1
1 2
2 1
2 2
4
Illustration for the first example:
Illustration for the second example:
题意：唔就是，让你找到一个最小的等腰三角形，使得给出的所有点都包含在等腰三角形里或者等腰三角形边上
分析: 其实就是找给出的点在y轴上截距最大的时候，满足方程y=-x+b,移一下就是,x+y=b，只要找到x+y的最大值即可、
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 #include<cmath>
5 using namespace std;
6 int main()
7 {
8 int n;
9 while(~scanf("%d",&n))
10 {
11 int a,b,ans=0;
12 while(n--)
13 {
14 scanf("%d %d",&a,&b);
15 ans=max(a+b,ans);
16 }
17 printf("%d\n",ans);
18 }
19 return 0;
20 }
C - C Karen and Morning
Description
Karen is getting ready for a new school day!
It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.
What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?
Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50.
Input
The first and only line of input contains a single string in the format hh:mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59).
Output
Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome.
Sample Input
05:39
11
13:31
0
23:59
1
Hint
In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.
In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.
In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome.
题意：这个女孩每到一个回文时间就会醒来，问她最少能睡多长时间。输入的字符串格式一定是hh:xx格式
分析：模拟就好
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 int main() 8 { 9 char s[10]; 10 while(~scanf("%s",s)) 11 { 12 int a,b; 13 a=(s[0]-‘0‘)*10+s[1]-‘0‘;//a是小时 14 b=(s[3]-‘0‘)*10+s[4]-‘0‘;//b是分钟 15 //printf("%d %d",a,b); 16 int q,w,e,r,j,t,tt,flag=0,i; 17 for(i=a;i<=23;i++)//注意只有1-23小时 18 { 19 q=i/10; 20 w=i%10; 21 if(i==a) 22 j=b; 23 else 24 j=0; 25 for(;j<=59;j++)//只有0-59秒 26 { 27 28 e=j/10; 29 r=j%10; 30 if(q==r&&w==e) 31 { 32 t=i,tt=j; 33 flag=1; 34 break; 35 } 36 } 37 if(flag==1) 38 break; 39 } 40 // printf("%d %d\n",t,tt); 41 if(flag==1) 42 { 43 if(t==a) 44 printf("%d\n",tt-b);//当在原来的小时形成回文时，只需要减去分钟数就好 45 else 46 printf("%d\n",(t-a-1)*60+60-b+tt);//间隔了t-a-1个小时 47 } 48 if(flag==0&&i==24) 49 printf("%d\n",(24-a-1)*60+60-b); 50 } 51 return 0; 52 }
D - D
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows n coffee recipes. The i-th recipe suggests that coffee should be brewed between li and ri degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least k recipes recommend it.
Karen has a rather fickle mind, and so she asks q questions. In each question, given that she only wants to prepare coffee with a temperature between a and b, inclusive, can you tell her how many admissible integer temperatures fall within the range?
The first line of input contains three integers, n, k (1 ≤ k ≤ n ≤ 200000), and q (1 ≤ q ≤ 200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next n lines describe the recipes. Specifically, the i-th line among these contains two integers li and ri (1 ≤ li ≤ ri ≤ 200000), describing that the i-th recipe suggests that the coffee be brewed between li and ri degrees, inclusive.
The next q lines describe the questions. Each of these lines contains a and b, (1 ≤ a ≤ b ≤ 200000), describing that she wants to know the number of admissible integer temperatures between a and b degrees, inclusive.
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between a and b degrees, inclusive.
3 2 4
91 94
92 97
97 99
92 94
93 97
95 96
90 100
3
3
0
4
2 1 1
1 1
200000 200000
90 100
0
In the first test case, Karen knows 3 recipes.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
题意：有n个食谱，告诉你n段温度适合煮咖啡，如果某个温度被至少k个食谱里的温度包含，则符合条件，你有q次查询，让你求出每次查询中符合条件的温度的数量
分析：看数据量，暴力肯定不行，看代码吧
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 const int m = 2e5; 5 int main() 6 { 7 int n,k,q,a[m]; 8 while(~scanf("%d %d %d",&n,&k,&q)) 9 { 10 int l,r; 11 memset(a,0,sizeof(a)); 12 for(int i=1;i<=n;i++) 13 { 14 scanf("%d %d",&l,&r); 15 a[l]++; 16 a[r+1]--; 17 } 18 for(int i=1;i<=m;i++) 19 { 20 a[i]+=a[i-1]; 21 } 22 for(int i=1;i<=m;i++) 23 { 24 a[i]=a[i-1]+(a[i]>=k); 25 } 26 for(int i=1;i<=q;i++) 27 { 28 scanf("%d %d",&l,&r); 29 printf("%d\n",a[r]-a[l-1]); 30 } 31 } 32 return 0; 33 }
不会解释，贴一个题解的链接https://blog.csdn.net/nka_kun/article/details/73411740
E - E Karen and Game 816c
On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.
Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!
The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).
If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.
The next k lines should each contain one of the following, describing the moves in the order they must be done:
If there are multiple optimal solutions, output any one of them.
3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1
4
row 1
row 1
col 4
row 3
3 3
0 0 0
0 1 0
0 0 0
-1
3 3
1 1 1
1 1 1
1 1 1
3
row 1
row 2
row 3
In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.
In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.
题意：你有一个n×m的矩阵，且全为0，要变成所给出的矩阵，操作要么每行都加1，要么每列都加1，如果可以输出操作的行和列，否则输出-1
分析：这道题其实就是一个模拟的过程，将所给出的矩阵每列或每行减一，看最后的矩阵是否全为0，但要注意两个问题：1：数组大小的问题，一开始数组开太小导致w了好多发；2：当行数小于列数时，先执行消除行的操作；当列数小于行数的时候，先执行消除列的操作。
#include<cstdio> #include<algorithm> #include<iostream> #include<cstring> #include<cmath> using namespace std; int main() { int a[505][505],b[505][505],c[505],d[50005],aa[50005],bb[50005],n,m,e[50005],f[50005]; while(~scanf("%d %d",&n,&m)) { memset(b,0,sizeof(b)); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { scanf("%d",&a[i][j]); } // printf("%d\n",c[i]); }//找到每一行最小的那个数 int t=0,tt=0; if(n<m)//如果行数小于列数，先消除行 { for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { d[j]=a[i][j]; } sort(d,d+m); c[i]=d[0]; } for(int i=0;i<n;i++) { for(int k=0;k<c[i];k++) aa[t++]=i; // printf("row %d\n",i); for(int j=0;j<m;j++) { a[i][j]-=c[i];//每一行每个数减去最小的那个数 } // for(int j=1;j<=m;j++) // { // printf("%d ",a[i][j]); // } // printf("\n"); } for(int j=0;j<m;j++) { for(int i=0;i<n;i++) { e[i]=a[i][j]; } sort(e,e+n); // printf("%d\n",e[0]); f[j]=e[0];//找到每列最小的那个数 } for(int j=0;j<m;j++) { for(int k=0;k<f[j];k++) bb[tt++]=j; // printf("col %d\n",j); for(int i=0;i<n;i++) { a[i][j]-=f[j];//每列每个数减去每列最小的那个数 } } } else//如果列数小于行数，先消除列数 { for(int j=0;j<m;j++) { for(int i=0;i<n;i++) { e[i]=a[i][j]; } sort(e,e+n); // printf("%d\n",e[0]); f[j]=e[0];//找到每列最小的那个数 } for(int j=0;j<m;j++) { for(int k=0;k<f[j];k++) bb[tt++]=j; // printf("col %d\n",j); for(int i=0;i<n;i++) { a[i][j]-=f[j];//每列每个数减去每列最小的那个数 } } for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { d[j]=a[i][j]; } sort(d,d+m); c[i]=d[0]; } for(int i=0;i<n;i++) { for(int k=0;k<c[i];k++) aa[t++]=i; // printf("row %d\n",i); for(int j=0;j<m;j++) { a[i][j]-=c[i];//每一行每个数减去最小的那个数 } // for(int j=1;j<=m;j++) // { // printf("%d ",a[i][j]); // } // printf("\n"); } } int flag=0; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(a[i][j]!=b[i][j])//b为全是0的矩阵 { flag=1; break; } } }//与全是0的矩阵比较，如果不相等则跳出 if(flag==1) { printf("-1\n"); } else { printf("%d\n",t+tt); for(int i=0;i<t;i++) printf("row %d\n",aa[i]+1); for(int i=0;i<tt;i++) printf("col %d\n",bb[i]+1); } } return 0; }
标签：ssis pie ever cer details coff sdi 字符串 erro
原文地址：https://www.cnblogs.com/LLLAIH/p/9742663.html
