A. Road To Zero
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given two integers x and y. You can perform two types of operations:
Pay a dollars and increase or decrease any of these integers by 1. For example, if x=0 and y=7 there are four possible outcomes after this operation:
x=0, y=6;
x=0, y=8;
x=−1, y=7;
x=1, y=7.
Pay b dollars and increase or decrease both integers by 1. For example, if x=0 and y=7 there are two possible outcomes after this operation:
x=−1, y=6;
x=1, y=8.
Your goal is to make both given integers equal zero simultaneously, i.e. x=y=0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1≤t≤100) — the number of testcases.
The first line of each test case contains two integers x and y (0≤x,y≤109).
The second line of each test case contains two integers a and b (1≤a,b≤109).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
inputCopy
2
1 3
391 555
0 0
9 4
outputCopy
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391+555+391=1337 dollars.
In the second test case both integers are equal to zero initially, so you dont’ have to spend money.
题意：
题意不算难懂，简单说下：
1.花费a元：对x，y其中
任意一个
增大或减小1
2.花费b元：对x，y
同时
增加或者减小1，即同增同减
求最少花多少元能让x，y都变成0
思路：
思路是找出两个数的差，这部分通过第一步来操作，剩下的部分由b来操作
但是具体要分两种情况
- x，y同号：
- x，y异号：
具体操作看代码，比较简单，不多说了
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
typedef long long ll;
int main()
{
int t;
ll x, y;
ll a, b;
cin >> t;
while (t--)
{
cin >> x >> y;
cin >> a >> b;
if (x == 0 && y == 0)
{
cout << '0' << endl;
continue;
}
else
{
ll m = max(x, y);
ll n = min(x, y);
ll u = m - n;
ll ans;
if (x * y >= 0)
ans = min((x + y) * a, n * b + u * a);
else
ans = min(n * b + (m + n) * a, u * a);
cout << ans << endl;
}
}
return 0;
}
B. Binary Period
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Let’s say string s has period k if si=si+k for all i from 1 to |s|−k (|s| means length of string s) and k is the minimum positive integer with this property.
Some examples of a period: for s=“0101” the period is k=2, for s=“0000” the period is k=1, for s=“010” the period is k=2, for s=“0011” the period is k=4.
You are given string t consisting only of 0’s and 1’s and you need to find such string s that:
String s consists only of 0’s and 1’s;
The length of s doesn’t exceed 2⋅|t|;
String t is a subsequence of string s;
String s has smallest possible period among all strings that meet conditions 1—3.
Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t=“011” is a subsequence of s=“10101”.
Input
The first line contains single integer T (1≤T≤100) — the number of test cases.
Next T lines contain test cases — one per line. Each line contains string t (1≤|t|≤100) consisting only of 0’s and 1’s.
Output
Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.
Example
inputCopy
4
00
01
111
110
outputCopy
00
01
11111
1010
Note
In the first and second test cases, s=t since it’s already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.
In the third test case, there are shorter optimal solutions, but it’s okay since we don’t need to minimize the string s. String s has period equal to 1.
题意
就是给你一个字符串t（只包含0和1），让你求周期最小的字符串s，并输出。
s满足的条件：
- 字符串s只包含0和1;
- s的长度不超过2⋅t ;（t是字符串t的长度）
- 字符串t是字符串s的子序列;
在满足条件1-3的所有字符串中，字符串s的周期可能最小。
思路：
题目中已经给了足够的提示，就是通过字符的增添和删减来构造新的字符串
-
很显然如果字符串t全由1或者0构成，此时周期就为
1
，不需要构建新的，直接输出t就可 -
但是既有1又有0时，容易想到最小周期是
2
，这就需要构造01010101…
或者10101010…这两种序列
具体实现请看代码
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
typedef long long ll;
int main()
{
int T;
char t[101], s[202];
cin >> T;
while (T--)
{
cin >> t;
int flag = 0;
int len = strlen(t);
int sum = 0;
for (int i = 0; i < len; i++)
sum += (t[i]-'0');
if ((sum == 0) || (sum == len))
{
cout << t << endl;
continue;
}
else
{
for (int i = 0; i < len-1; i++)
{
if (t[i] == t[i + 1] && t[i] == '0')
{
cout << t[i] << '1'; continue;
}
else if (t[i] == t[i + 1] && t[i] == '1')
{
cout << t[i] << '0'; continue;
}
else
{
cout << t[i]; continue;
}
}
cout << t[len - 1] << endl;
}
}
return 0;
}
比赛时就做出了AB两道题，c题的题意完全不懂，比赛结束后才弄懂的
C. Yet Another Counting Problem
time limit per test3.5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given two integers a and b, and q queries. The i-th query consists of two numbers li and ri, and the answer to it is the number of integers x such that li≤x≤ri, and ((xmoda)modb)≠((xmodb)moda). Calculate the answer for each query.
Recall that ymodz is the remainder of the division of y by z. For example, 5mod3=2, 7mod8=7, 9mod4=1, 9mod9=0.
Input
The first line contains one integer t (1≤t≤100) — the number of test cases. Then the test cases follow.
The first line of each test case contains three integers a, b and q (1≤a,b≤200; 1≤q≤500).
Then q lines follow, each containing two integers li and ri (1≤li≤ri≤1018) for the corresponding query.
Output
For each test case, print q integers — the answers to the queries of this test case in the order they appear.
Example
inputCopy
2
4 6 5
1 1
1 3
1 5
1 7
1 9
7 10 2
7 8
100 200
outputCopy
0 0 0 2 4
0 91
题意：
[l,r][l,r][l,r]范围内多少个数满足 (x % b) % a != (x % a) % b。
思路：
打表找规律。
假设b>ab > ab>a
可以发现当为 lcm(a,b)倍数的时候，再连续b个数字，是那个式子满足相等的。
找出多少个相等的，就可以知道不相等的个数了。
明显对于k∗lcm(a,b)的数，%b或者%a都为0。那么肯定相等。连续b个数字下，模b不变，只用看模a，所以肯定相等。
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
ll a, b, q;
ll Lcm;
ll gcd(ll n, ll m) {
return m == 0 ? n : gcd(m, n % m);
}
ll lcm(ll n, ll m) {
return n / gcd(n, m) * m;
}
ll cal(ll x) {
ll res = 0;
ll num = x / Lcm;
res += num * b;
if (num != 0) {
x %= num * Lcm;
}
x += 1;
res += min(x, b);
return res;
}
int main() {
int t;
scanf("%d", &t);
while (t--)
{
scanf("%lld%lld%lld", &a, &b, &q);
if (a > b) swap(a, b);
Lcm = lcm(a, b);
while (q--) {
ll l, r; scanf("%lld%lld", &l, &r);
ll ans = cal(r) - cal(l - 1);
ans = r - l + 1 - ans;
printf("%lld ", ans);
}
printf("\n");
}
return 0;
}
本人水平有限，如有错误，请指正