USC 1329 Decode 坑坑坑gets

其实一贯都是用string来着。。

其实后来看刘汝佳的书又喜欢上了fgets然后一路过关斩将来着。。

这次觉得题水就用了gets啊啊啊啊

然后就WA了9次啊有木有！！！

H. Decode TimeLimit: 1s   MemoryLimit: 64M Description Bruce Force has had an interesting idea how to encode strings. The following is the description of how the encoding is done: Let x 1 ,x 2 ,...,x n  be the sequence of characters of the string to be encoded. 1.Choose an integer  m and  n pairwise distinct numbers p 1,p 2,...,p n from the set { 1,  2, ...,  n} (a permutation of the numbers  1 to  n). 2.Repeat the following step  m times. 3.For  1 ≤ i ≤  n set y i to x pi, and then for  1 ≤ i ≤  n replace x i by y i. For example, when we want to encode the string "hello", and we choose the value  m = 3  and the permutation  2, 3, 1, 5, 4 , the data would be encoded in 3 steps: "hello" - > "elhol" -> "lhelo" -> "helol". Bruce gives you the encoded strings, and the numbers  m  and p 1 , ..., p n  used to encode these strings. He claims that because he used huge numbers  m  for encoding, you will need a lot of time to decode the strings. Can you disprove this claim by quickly decoding the strings? Input The input contains several test cases. Each test case starts with a line containing two numbers nand m (1 ≤ n ≤ 80, 1 ≤ m ≤ 109). The following line consists of n pairwise different numbers p1,...,pn (1 ≤ pi ≤ n). The third line of each test case consists of exactly n characters, and represent the encoded string. The last test case is followed by a line containing two zeros. Output For each test case, print one line with the decoded string. SampleInput 5 3 2 3 1 5 4 helol 16 804289384 13 10 2 7 8 1 16 12 15 6 5 14 3 4 11 9 scssoet tcaede n 8 12 5 3 4 2 1 8 6 7 encoded? 0 0 SampleOutput hello second test case encoded?

不说啥了，就是mod一个周期就行。getline的代码在最下面，纪念一下死的多坑。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#include 
#include 
#include 
#include 
using namespace std;
const int MAXN = 888;
int n, m;
int a[MAXN], f[MAXN];
char ans[MAXN], s[MAXN], ch;

int to(int v, int w){
    while( w-- )    v = a[v];
    return v;
}

int main(){    
    while( EOF != scanf("%d %d\n", &n, &m) ){
        if( !n && !m )  break;
        //getchar();
        for(int i=1; i<=n; i++)
            cin>>a[i];//scanf("%d", &a[i]), getchar();
        cin.getline(s+1, MAXN);
        cin.getline(s+1, MAXN);
        for(int i=1; i<=n; i++)
            if( a[i]==i )   f[i] = 1;
            else{
                int t=a[i], cnt=1;
                while( t!=i )    t=a[t],     cnt++;
                f[i] = cnt;
            }
        for(int i=1; i<=n; i++)
            ans[ to(i, m%f[i]) ] = s[i];
        ans[n+1] = 0;
        cout<

真切感慨以后在时间充裕的时候能用C++就用C++。别自作多情的装逼用c啥啥的