S1
、S2
,问是否能找出对两字串皆合法的 L
?N
代表测资数S1
、S2
皆为二进位字串L
?
S - L
(二进位减法) 直到 S = L
p
代表第几笔测资S - L
,其实就是看 S
是不是 L
的倍数,题目有举例:S = 11011 = 27
L = 11 = 3
bin dec
11011 = 27
11000 = 24
10101 = 21
10010 = 18
1111 = 15
1100 = 12
1001 = 9
110 = 6
11 = 3
S1
、S2
都合法,代表 L
要同时为两者的公因数,所以用 GCD 检查两者是否互质 (GCD = 1);先把输入转成十进位比较好处理N
後,用 while
回圈重复读入两字串存到字元阵列里
int N;
scanf("%d", &N);
while(N--){
char S1[31] = {0};
char S2[31] = {0};
scanf("%s%s", S1, S2);
...
}
int str_to_int(char *a, char *b){
int num1 = 0, num2 = 0;
int i;
for(i = 0; i < strlen(a); i++){
num1 = (num1 << 1) + (a[i] - '0');
}
for(i = 0; i < strlen(b); i++){
num2 = (num2 << 1) + (b[i] - '0');
}
return GCD(num1, num2);
}
int GCD(int a, int b){
return b == 0 ? a : GCD(b, a % b);
}
if(str_to_int(S1, S2) != 1){
printf("Pair #%d: All you need is love!\n", Case++);
}
else{
printf("Pair #%d: Love is not all you need!\n", Case++);
}
#include<stdio.h>
#include<string.h>
int GCD(int a, int b){
return b == 0 ? a : GCD(b, a % b);
}
int str_to_int(char *a, char *b){
int num1 = 0, num2 = 0;
int i;
for(i = 0; i < strlen(a); i++){
num1 = (num1 << 1) + (a[i] - '0');
}
for(i = 0; i < strlen(b); i++){
num2 = (num2 << 1) + (b[i] - '0');
}
return GCD(num1, num2);
}
int main(){
int N, Case = 1;
scanf("%d", &N);
while(N--){
char S1[31] = {0};
char S2[31] = {0};
scanf("%s%s", S1, S2);
if(str_to_int(S1, S2) != 1){
printf("Pair #%d: All you need is love!\n", Case++);
}
else{
printf("Pair #%d: Love is not all you need!\n", Case++);
}
}
return 0;
}
stoi()
dec
、hex
、oct
可用
gcd()
,但 C++17 才支援
#include <bits/stdc++.h>
using namespace std;
int GCD(int a, int b){
return b == 0 ? a : GCD(b, a % b);
}
int main(){
int N, Case = 1;
cin >> N;
while(N--){
string S1, S2;
cin >> S1 >> S2;
int num1 = stoi(S1, nullptr, 2), num2 = stoi(S2, nullptr, 2);
if(GCD(num1, num2) != 1){
cout << "Pair #" << Case++ << ": All you need is love!\n";
}
else{
cout << "Pair #" << Case++ << ": Love is not all you need!\n";
}
}
return 0;
}
要撰写前端功能,直接使用JavaScript是绝对可行的,但要更有效率、具有良好开发体验的话,使用L...
到今天为止介绍了不少应用於k8s上的服务,并且大部分都可以透过operator的方式进行同性质的服务...
透过 WordPress 区块编辑器撰写文章最常用的「区块 Block」之一,就是「标题区块 He...
铁人赛的前20天,把常常混淆的那些观念重新理解过一遍之後,後面的10天打算来练习切版,希望透过实际练...
运算式树 Expression tree Expression tree 是一个树状结构的物件, 这...