标签：ring += ack org closed fine long size splay
这道题就是区间加一个等差数列，然后最后求每一个数的值。
O(n)做法：二阶差分。
其实就是差分两遍。举个例子 0 0 0 0 0 0 0，变成了 0 2 4 6 8 0 0。第一遍差分：0 2 2 2 2 -8 0，然后在这个差分数组上在进行差分，得到 0 2 0 0 0 10 8，完事。
我们在用通项搞一遍：等差数列的首项是s，末项是e，公差是d，则
最终的样子：0 s s+d s+2d e 0 0，
第一遍差分：0 s d d d -e 0，
第二遍差分：0 s d-s 0 0 -e-d e 。
所以我们只用维护四个值就行了。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(‘ ‘) 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 1e7 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ‘ ‘; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - ‘0‘; ch = getchar();} 27 if(last == ‘-‘) ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar(‘-‘); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + ‘0‘); 35 } 36 37 int n, m; 38 ll dif[maxn], sum = 0, Max = -INF; 39 40 int main() 41 { 42 n = read(); m = read(); 43 for(int i = 1; i <= m; ++i) 44 { 45 int L = read(), R = read(), s = read(), e = read(); 46 int d = (e - s) / (R - L); 47 dif[L] += s; dif[L + 1] += d - s; 48 dif[R + 1] -= e + d; dif[R + 2] += e; 49 } 50 for(int i = 1; i <= n; ++i) dif[i] = dif[i - 1] + dif[i]; 51 for(int i = 1; i <= n; ++i) 52 { 53 dif[i] = dif[i - 1] + dif[i]; 54 sum ^= dif[i]; 55 Max = max(Max, dif[i]); 56 } 57 write(sum); space; write(Max); enter; 58 return 0; 59 }
标签：ring += ack org closed fine long size splay
原文地址：https://www.cnblogs.com/mrclr/p/9744883.html
