#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
int a,b;
while(scanf("%d/%d",&a,&b)!=EOF)
{
int yushu[10000],ans[10000];
int j=0,k=0;
int jdg=0;
while(a%b)
{
yushu[k++]=a%b;
ans[j++]=a/b;
a=a%b*10;
for(int i=0;i<k-1;i++)
{
if(yushu[i]==yushu[k-1])
{
for(int p=i+1;p<=k-1;p++)
cout<<ans[p];
cout<<endl;
jdg=1;
break;
}
}
if(jdg==1)
break;
}
if(a%b==0)
cout<<0<<endl;
}
return 0;
}
2895. 循环小数
Saitama
10165101122
用循环会超时，所以改用了标记
include
int main()
{
int a,b;
while(scanf(“%d/%d”,&a,&b)!=EOF)
{ if(a<0||b<0)break;
int i=0,k,flag=0;
int s[100000]={0};
int yu[100000]={0};
for(;a>=b;a=a%b);
for(;a<b;++i)
{ if(a==0)
{
printf(“0\n”);
break;
}
if(yu[a]==1)
{
for(k=0;k<i;++k)
{
if(s[k]==(a10)/b)break;
}
for(;k<i;++k)printf(“%d”,s[k]);
printf(“\n”); flag=1;
break;
}
yu[a]=1;
a=a10;
s[i]=a/b;
a=a%b;
if(flag==1)break;
}
}
}
Money4
判断余数是否出现过
用map标记会超时
#include<bits/stdc++.h>
using namespace std;
int main() {
char c;
int m, n, i;
while (scanf("%d/%d", &m, &n) != EOF) {
int nums[100001] = {0};
bool flag = false;
string res = {}, ans;
i = 1;
nums[m] = i++;
int left;
while (m != 0) {
m *= 10;
left = m / n;
res += to_string(left);
m = m % n;
if (nums[m]) {
for (int j = nums[m]-1; j < i-1; j++)
printf("%c",res[j]);
printf("\n");
break;
}
else {
nums[m] = i++;
}
}
if (m==0) printf("0\n");
}
}
296804129
求助，帮忙看看哪里错了
include
include
int main()
{
int m,n;
int a[1000],b[1000],i,j;
while(scanf(“%d/%d”,&m,&n)==2)
{
if(m<=0||n<=0)break;
m=m10;
for(i=0; m%n!=0; i++)
{
a[i]=m/n;
b[i]=m%n;
m=b[i]10;
for(j=0; j<i; j++)
{
if(b[i]==b[j]&&a[j]==a[i])break;
}
if(i!=0&&a[j]==a[i]&&j!=i)break;
}
if(m%n==0)printf(“0\n”);
else
{
for(; j<i; j++)printf(“%d”,a[j]);
printf(“\n”);
}
}
return 0;
}
一剑无痕雪满山
烦烦烦 总是不过
[em:15]我也……而且就是按照老师的方法。。总是tle 。。。
_(:зゝ∠)_哭了
include
include
include
int main()
{
int n,m;int i;int flag;int yushu;int shang;
while (scanf(“%d/%d”,&m,&n))
{
if (m<=0||n<=0) {break;}
yushu=(int)malloc(nsizeof(int));
shang=(int)malloc(nsizeof(int));
for (i=0;i<n;i++)
{
yushu[i]=-1;shang[i]=-1;
}
for (i=0;i<n;i++)
{
yushu[m]=i;
m=m*10;
shang[i]=m/n;
m=m%n;
if (m==0) {printf("0\n");break;}
if (yushu[m]!=-1)
{
for(flag=yushu[m];flag<=i;flag++)
{
printf("%d",shang[flag]);
}
printf("\n");break;
}
}
free(yushu);
free(shang);
}
return 0;
}
Suzuki_Yuuta
坑点就是寻找当前的余数是否与之前余数相同的操作时需要优化，否则会超时。
void solve(int a, int b) {
int quotient[100001], i_q = 1; \\ 记录商以及个数
int remain[100001] = {0}, i_r = 1; \\ 用类似哈希表的方式存储余数及其对应的次序
int p = a; bool find = false;
while (!find) {
quotient[i_q++] = p / b;
if (p % b == 0) {
cout << 0 << endl; find = true;
} else {
if (remain[p % b]) {
for (int j = remain[p % b] + 1; j < i_q; j++) {
cout << quotient[j];
}
cout << endl; find = true;
}
remain[p % b] = i_r++; p = (p % b) * 10;
}
}
}
LzQuarter
汗，建议确认过了样例以后试试
https://blog.csdn.net/panhe1992/article/details/7471320
上面这个地址提到的样例
#include<iostream>
#include<queue>
using namespace std;
int main(){
int a, b;
char t;
while(cin >> a >> t >> b){
int vis[100001] = {0};
queue<int> left;
queue<int> res;
int zero = 0;
do{
int resnum = a / b;
int leftnum = a % b;
//cout << resnum << " " << leftnum << endl;
a %= b;
if(leftnum == 0){
zero = 1;
break;
}
if(vis[leftnum]){
int firstpoped = 0;
while(left.front() != leftnum){
firstpoped = 1;
left.pop();
res.pop();
}
if(res.front() != resnum){
left.pop();
res.pop();
res.push(resnum);
}
else if(!firstpoped && res.front() == 0){
left.pop();
res.pop();
res.push(resnum);
}
break;
}
else{
vis[leftnum] = 1;
left.push(leftnum);
res.push(resnum);
}
a *= 10;
}while(1);
if(zero){
cout << "0" << endl;
}
else{
while(!res.empty()){
cout << res.front();
res.pop();
}
cout << endl;
}
}
return 0;
}
用unordered_map可以