选用 BKDRhash 方式
参考资料：https://www.byvoid.com/blog/string-hash-compare/
字符串通篇资料：https://blog.csdn.net/ck_boss/article/details/47066727
HDU 4821 string
传送门：https://vjudge.net/problem/HDU-4821
题意：给出M和L，和一个字符串S。要求找出S的子串中长度为L*M，并且可以分成M段，每段长L，并且M段都不相同的子串个数。
思路：一道字符串哈希题。哈希的方法是BKDRHash，哈希中进制是31,131等素数。
#include <iostream> #include <map> #include <string> #include <cstring> #include <cstdio> #define maxn 100005 #define SEED 31 #define ULL unsigned long long using namespace std; ULL k[maxn],ha[maxn]; map <ULL , int> mp; int m,l; char s[maxn]; int main(){ while (scanf("%d%d",&m,&l)!=EOF){ scanf("%s",s); k[0]=1; int len = strlen(s); ha[len] = 0; for (int i=1;i<=l;i++) k[i] = k[i-1]*SEED; for (int i=len-1;i>=0;i--) ha[i] = ha[i+1]*SEED + s[i] - 'a' + 1; int ans=0; for (int i=0;i<l && i+m*l<len;i++){ mp.clear(); for (int j=i;j<i+m*l;j+=l) mp[ha[j]-ha[j+l]*k[l]]++; if (mp.size() == m) ans++; for (int j=i+m*l;j<=len-l;j+=l){ int head = j-m*l; ULL x = ha[head] - ha[head+l]*k[l]; mp[x]--; if (mp[x] == 0) mp.erase(x); mp[ha[j] - ha[j+l]*k[l]]++; if (mp.size() == m) ans ++; } } cout << ans << endl; } } HDU 4080 Stammering Aliens
题意
每组数据为一个整数m和一个长度不小于m的字符串，求该字符串的一个子串，该子串在满足出现次数不小于m的同时应尽量长。
输出该长度和最右侧出现的起始位置。如果存在多组数据，输出有最靠近右侧的那组。
m<=4e5,时限5s
字符串hash+二分答案。我写的版本被卡了，从网上找了一个相同写法的4000ms卡过去了，学习一下。
by Vicente:
#include <cstring> #include <iostream> #include <map> #define maxn 40005 #define SEED 31 #define ULL unsigned long long using namespace std; int m,len; ULL k[maxn],ha[maxn]; char s[maxn]; map <ULL , int> mp; bool isOK(int L){ mp.clear(); for (int i=0;i+L<=len;i++){ mp[ha[i]-ha[i+L]*k[L]]++; if (mp[ha[i] - ha[i+L]*k[L]]>=m)return true; } return false; } int main(){ while (scanf("%d",&m) && m!=0){ scanf("%s",s); len = strlen(s); k[0] = 1; for (int i=1;i<=len;i++) k[i] = k[i-1] * SEED; ha[len] = 0; for (int i = len-1;i>=0;i--) ha[i] = ha[i+1] * SEED + s[i] - 'a'+1; int l=m,r=len,mid; while (l<r){ mid = (l+r+1) >>1; if (isOK(mid)) l = mid;else r=mid-1; } if (!isOK(r)) { puts("none"); continue; }else{ cout << r << " "; mp.clear(); for (int i=0;i+r<=len;i++){ mp[ha[i]-ha[i+r]*k[r]]++; if (mp[ha[i] - ha[i+r]*k[r]]>=m) cout << i << endl; } } } } AC版本：
#include <cstring> #include <iostream> #include <map> #define ull unsigned long long using namespace std; const int maxn = (int)4e4 + 10; int n,len; char s[maxn]; ull has[maxn]; ull p = 2333,a[maxn]; bool judge(int x) { map<int,int> mp; ull res; for (int i = 1;i + x - 1 <= len;i ++) { res = has[i + x - 1] - has[i - 1] * a[x]; if (++mp[res] >= n) return 1; } return 0; } int check(int x) { map<int,int> mp; int pos = -1; ull res; for (int i = 1;i + x - 1 <= len;i ++) { res = has[i + x - 1] - has[i - 1] * a[x]; if (++mp[res] >= n) pos = i; } return pos - 1; } int main() { while (~scanf("%d",&n) && n) { scanf("%s",s + 1); len = strlen(s + 1); has[0] = 0,a[0] = 1; for (int i = 1;i <= len;i ++) { has[i] = has[i - 1] * p + s[i]; a[i] = a[i - 1] * p; } int l = 1,r = len + 1,mid; while (l <= r) { mid = (l + r) >> 1; if (judge(mid)) l = mid + 1; else r = mid - 1; } if (r == 0) printf("none\n"); else printf("%d %d\n",r,check(r)); } return 0; } 哈哈哈经过学习，我改了一点内容，终于也把这道题卡过去了。
4773ms…说是卡过去真是一点不错。
AC代码 by Vicente.
#include <cstring> #include <iostream> #include <map> #define maxn 40005 #define SEED 31 #define ULL unsigned long long using namespace std; int m,len; ULL k[maxn],ha[maxn]; char s[maxn]; map <ULL , int> mp; bool isOK(int L){ mp.clear(); ULL res; for (int i=0;i+L<=len;i++){ res=ha[i]-ha[i+L]*k[L]; if (++mp[res]>=m)return true; } return false; } int main(){ while (scanf("%d",&m) && m!=0){ scanf("%s",s); len = strlen(s); k[0] = 1; ha[len] = 0; for (int i=1;i<=len;i++)//强行放在一起写，就没爆T了 k[i] = k[i-1] * SEED,ha[len-i] = ha[len-i+1] * SEED + s[len-i] - 'a'+1;; // for (int i = len-1;i>=0;i--) ha[i] = ha[i+1] * SEED + s[i] - 'a'+1; int l=1,r=len+1,mid; while (l<=r){ mid = (l+r) >>1; if (isOK(mid)) l = mid+1;else r=mid-1; } if (r == 0) { puts("none"); continue; }else{ cout << r << " "; mp.clear(); int x; for (int i=0;i+r<=len;i++){ mp[ha[i]-ha[i+r]*k[r]]++; if (mp[ha[i] - ha[i+r]*k[r]]>=m) x=i;//cout << i << endl; } cout << x << endl;//不用x标记的话会WA，原因自己想 } } } 文章来源: https://blog.csdn.net/qq_42447015/article/details/97242753