POJ 1200 Crazy Search(RK)

题意

给定一个由NC个字母组成的字符串，求长度为N的不同子串的个数

思路：

由于只有NC个字母，可以将字母编号，0 ～ NC - 1，转换成数字，就可以将字符串表示成NC进制的数字，这样所有字串代表的数字都是唯一的，转换成10进制的数也是唯一的！

就像10的二进制表示只有1010

例如 

3 4

daababac

d = 3

a = 0

b = 1

c = 2

daa = 3 * 4 ^ 2 + 0 * 4 ^ 1 + 0 * 4 ^ 0 = 48

//vs1.0

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#define MAX 16000000



char str[MAX];

bool hash[MAX];

int ancii[128];



int main() {

    int N, NC;

    while (scanf("%d%d%s", &N, &NC, str) != EOF) {

        memset(hash, true, sizeof(hash));

        memset(ancii, 0, sizeof(ancii));

        int cnt = 0;

        for (char *s=str; *s; ++s) {

            if (!ancii[*s]) {

                ancii[*s] = ++cnt;

                if (NC == cnt) break;

            }

        }

        int sum;

        int ans = 0;

        int len = strlen(str) - N + 1;

        for (int i=0; i<len; ++i) {

            sum = 0;

            for (int j=0; j<N; ++j) sum = sum * NC + (ancii[str[i+j]] - 1);

            if (hash[sum]) {

                ++ans;

                hash[sum] = false;

            }

        }

        printf ("%d\n", ans);

    }

    return 0;

}







//vs2.0

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#define MAX 16000000



char str[MAX];

bool hash[MAX];

int ancii[128];



int main() {

    int N, NC;

    while (scanf("%d%d%s", &N, &NC, str) != EOF) {

        memset(hash, true, sizeof(hash));

        memset(ancii, 0, sizeof(ancii));

        int cnt = 0;

        for (char *s=str; *s; ++s) {

            if (!ancii[*s]) {

                ancii[*s] = ++cnt;

                if (NC == cnt) break;

            }

        }

        int sum = 0;

        int tmp = 1;

        for (int i=0; i<N; ++i) {

            tmp *= NC;

            sum = sum * NC + ancii[str[i]] - 1; 

        }

        tmp /= NC;

        hash[sum] = false;

        int ans = 1;

        int len = strlen(str);

        for (int i=N; i<len; ++i) {

            sum = (sum - tmp * (ancii[str[i-N]]-1)) * NC + ancii[str[i]] - 1;

            if (hash[sum]) {

                ++ans;

                hash[sum] = false;

            }

        }

        printf ("%d\n", ans);

    }

    return 0;

}