E. Kefa and Watch
Time Limit: 1 Sec ?
Memory Limit: 256 MB
題目連接
http://codeforces.com/contest/580/problem/EDescription
One day Kefa the parrot was walking down the street as he was on the way home from the restaurant when he saw something glittering by the road. As he came nearer he understood that it was a watch. He decided to take it to the pawnbroker to earn some money.
The pawnbroker said that each watch contains a serial number represented by a string of digits from?0?to?9, and the more quality checks this number passes, the higher is the value of the watch. The check is defined by three positive integers?l,?r?and?d. The watches pass a check if a substring of the serial number from?l?to?r?has period?d. Sometimes the pawnbroker gets distracted and Kefa changes in some substring of the serial number all digits to?c?in order to increase profit from the watch.
The seller has a lot of things to do to begin with and with Kefa messing about, he gave you a task: to write a program that determines the value of the watch.
Let us remind you that number?x?is called a period of string?s?(1?≤?x?≤?|s|), if?si??=??si?+?x?for all?i?from 1 to?|s|??-??x.
Input
The first line of the input contains three positive integers?n,?m?and?k?(1?≤?n?≤?105,?1?≤?m?+?k?≤?105) — the length of the serial number, the number of change made by Kefa and the number of quality checks.
The second line contains a serial number consisting of?n?digits.
Then?m?+?k?lines follow, containing either checks or changes.
The changes are given as 1?l?r?с?(1?≤?l?≤?r?≤?n,?0?≤?c?≤?9). That means that Kefa changed all the digits from the?l-th to the?r-th to be?c.
The checks are given as 2?l?r?d?(1?≤?l?≤?r?≤?n,?1?≤?d?≤?r?-?l?+?1).
Output
For each check on a single line print "YES" if the watch passed it, otherwise print "NO".
Sample Input
3 1 2
112
2 2 3 1
1 1 3 8
2 1 2 1
Sample Output
NO
YES
HINT
?
題意
給你一個字符集只有10的串 ，然后又兩個操作
1.區間更新
2.查詢lr區間的字符的周期是否為d
題解：
直接暴力線段樹hash就好了
查詢操作只有判斷(l,l+len-d)和(l+len-d+1,r)是否一樣就好了
可以用類似錯位來解釋
代碼來自peterpan
代碼：
//#pragma comment(linker, "/STACK:102400000,102400000") #include<cmath> #include<cstdio> #include<iostream> #include<algorithm> #include<cstdlib> #include<queue> #include<cstring> using namespace std; typedef long long ll; #define input freopen("/Users/peteryuanpan/data.txt","r",stdin)#define N 100010 int n, m, k, lens; char s[N];#define lch id<<1 #define rch id<<1|1 class HashTree{ public:int mod, p;ll powp[N], sump[N];void getpowp(){powp[0] = 1;sump[0] = 1;for(int i = 1; i < N; i++){powp[i] = powp[i-1] * p;if(powp[i] >= mod) powp[i] %= mod;sump[i] = sump[i-1] + powp[i];if(sump[i] >= mod) sump[i] %= mod;}}ll v[N << 2];int len[N << 2];char lazy[N << 2];void plant(int id,int l,int r){lazy[id] = '\0';if(l == r){v[id] = s[l];len[id] = 1;return;}int mid = (l + r) >> 1;plant(lch, l, mid);plant(rch, mid + 1, r);len[id] = len[lch] + len[rch];v[id] = v[lch] * powp[len[rch]] + v[rch];if(v[id] >= mod) v[id] %= mod;}void pushdown(int id){if(lazy[id] != '\0'){lazy[lch] = lazy[rch] = lazy[id];v[lch] = lazy[id] * sump[len[lch] - 1];if(v[lch] >= mod) v[lch] %= mod;v[rch] = lazy[id] * sump[len[rch] - 1];if(v[rch] >= mod) v[rch] %= mod;lazy[id] = '\0';}}void update(int id,int ql,int qr,int l,int r,char c){if(ql == l && qr == r){lazy[id] = c;v[id] = c * sump[len[id] - 1];if(v[id] >= mod) v[id] %= mod;return;}pushdown(id);int mid = (l + r) >> 1;if(qr <= mid) update(lch, ql, qr, l, mid, c);else if(mid < ql) update(rch, ql, qr, mid + 1, r, c);else update(lch, ql, mid, l, mid, c), update(rch, mid + 1, qr, mid + 1, r, c);v[id] = v[lch] * powp[len[rch]] + v[rch];if(v[id] >= mod) v[id] %= mod;}ll query(int id,int ql,int qr,int l,int r){if(ql == l && qr == r){return v[id];}pushdown(id);int mid = (l + r) >> 1;if(qr <= mid) return query(lch, ql, qr, l, mid);else if(mid < ql) return query(rch, ql, qr, mid + 1, r);else{ll t1 = query(lch, ql, mid, l, mid);ll t2 = query(rch, mid + 1, qr, mid + 1, r);ll t = t1 * powp[qr - (mid + 1) + 1] + t2;if(t >= mod) t %= mod;return t;}} }tree1, tree2;bool equal(int l1, int r1, int l2, int r2){if(tree1.query(1, l1, r1, 0, lens - 1) != tree1.query(1, l2, r2, 0, lens - 1)) return false;if(tree2.query(1, l1, r1, 0, lens - 1) != tree2.query(1, l2, r2, 0, lens - 1)) return false;return true; }bool judge(int l, int r, int d){if(r - l + 1 == d) return true;int l2 = l + d, r2 = r;int len = r2 - l2 + 1;int l1 = l, r1 = l1 + len - 1;return equal(l1, r1, l2, r2); }int main(){//input;tree1.mod = 1e9 + 7, tree1.p = 799817, tree1.getpowp();tree2.mod = 1e9 + 9, tree2.p = 451309, tree2.getpowp();scanf("%d%d%d",&n,&m,&k);scanf("%s",s);lens = (int)strlen(s);tree1.plant(1, 0, lens - 1), tree2.plant(1, 0, lens - 1);for(int i = 1; i <= m + k; i++){int ty, l, r;scanf("%d%d%d",&ty,&l,&r);l--, r--;if(ty == 1){char c[5];scanf("%s",c);tree1.update(1, l, r, 0, lens - 1, c[0]);tree2.update(1, l, r, 0, lens - 1, c[0]);}else{int d;scanf("%d",&d);printf("%s\n", judge(l, r, d) ? "YES" : "NO");}}return 0; }
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