题意：求字符串s的所有前缀出现次数之和。
http://www.cnblogs.com/jklongint/p/4446117.html
思路：用kmp做，简单且效率高。以前缀结尾的位置分类，令dp[i]为以结尾位置在i的前缀数量，那么dp[i] = cnt(j)(j~i是前缀)，而由kmp的next函数的转移性质，可得如下递推方程:dp[i] = dp[next[i]] + 1，把这个递推式不断展开，也就是i = next[i]不断迭代，那么+1的个数就是dp[i] = cnt(j)(j~i是前缀)这个里面的cnt(j)了。这就是next数组的性质，实在巧妙！
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d ", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef double db; 50 typedef long long LL; 51 typedef pair<int, int> pii; 52 typedef multiset<int> msi; 53 typedef set<int> si; 54 typedef vector<int> vi; 55 typedef map<int, int> mii; 56 57 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 58 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 59 const int maxn = 2e5 + 7; 60 const int md = 10007; 61 const int inf = 1e9 + 7; 62 const LL inf_L = 1e18 + 7; 63 const double pi = acos(-1.0); 64 const double eps = 1e-6; 65 66 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 67 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 68 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 69 template<class T>T condition(bool f, T a, T b){return f?a:b;} 70 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 71 int make_id(int x, int y, int n) { return x * n + y; } 72 73 struct KMP { 74 int next[1000010]; 75 void GetNext(char s[]) { 76 mem0(next); 77 next[0] = next[1] = 0; 78 for(int i = 1; s[i]; i++) { 79 int j = next[i]; 80 while(j && s[i] != s[j]) j = next[j]; 81 next[i + 1] = s[j] == s[i]? j + 1 : 0; 82 } 83 } 84 }; 85 86 KMP kmp; 87 int dp[maxn]; 88 char s[maxn]; 89 90 int main() { 91 //freopen("in.txt", "r", stdin); 92 int T, n; 93 cin >> T; 94 while (T--) { 95 scanf("%d%s", &n, s); 96 kmp.GetNext(s); 97 mem0(dp); 98 int ans = n; 99 rep_up1(i, n) { 100 if (kmp.next[i] == 0) continue; 101 dp[i] = dp[kmp.next[i]] + 1; 102 dp[i] %= md; 103 ans += dp[i]; 104 ans %= md; 105 } 106 cout << ans << endl; 107 } 108 return 0; 109 }