HDU - 1358 Period KMP循环节出现次数
2017-11-02 17:03
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跟上篇博文一样的方法,
可以先看上一篇,一样的方法
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define PI acos(-1.0)
#define in freopen("in.txt", "r", stdin)
#define out freopen("out.txt", "w", stdout)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 7, maxd = 1e4 + 7, mod = 1e9 + 7;
const ll INF = 0x7f7
4000
f7f7f;
int n, ans;
char s[maxn];
int f[maxn];
void init() {
f[0] = f[1] = 0;
for(int i = 1; i < n; ++i) {
int j = f[i];
while( j && s[i] != s[j]) j = f[j];
f[i+1] = (s[i] == s[j] ? j+1 : 0);
}
//for(int i = 0; i <= len; ++i)
// cout << f[i] << " ";
}
void solve() {
for(int i = 2; i <= n; ++i) {
int t = (i / (i-f[i]));
if( i % (i-f[i]) == 0 && t != 1) {
printf("%d %d\n", i, t);
}
}
puts("");
}
int main() {
int kase = 1;
while(~scanf("%d", &n) && n) {
scanf("%s", s);
init();
printf("Test case #%d\n", kase++);
solve();
}
return 0;
}
可以先看上一篇,一样的方法
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define PI acos(-1.0)
#define in freopen("in.txt", "r", stdin)
#define out freopen("out.txt", "w", stdout)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e6 + 7, maxd = 1e4 + 7, mod = 1e9 + 7;
const ll INF = 0x7f7
4000
f7f7f;
int n, ans;
char s[maxn];
int f[maxn];
void init() {
f[0] = f[1] = 0;
for(int i = 1; i < n; ++i) {
int j = f[i];
while( j && s[i] != s[j]) j = f[j];
f[i+1] = (s[i] == s[j] ? j+1 : 0);
}
//for(int i = 0; i <= len; ++i)
// cout << f[i] << " ";
}
void solve() {
for(int i = 2; i <= n; ++i) {
int t = (i / (i-f[i]));
if( i % (i-f[i]) == 0 && t != 1) {
printf("%d %d\n", i, t);
}
}
puts("");
}
int main() {
int kase = 1;
while(~scanf("%d", &n) && n) {
scanf("%s", s);
init();
printf("Test case #%d\n", kase++);
solve();
}
return 0;
}
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