http://acm.hdu.edu.cn/showproblem.php?pid=4893
题意:10万的区间,初始都为0,10万次操作,三种操作为单点修改,区间将每个数改成最近的斐波那契数,以及区间求和。
分析:用一个flag记录该段是否被改成斐波那契数,同时多维护一个sum1表示如果该段改成斐波那契数,区间和为多少。开始sum1为区间长度,之后在单点做了修改以后对其更新,需要的时候用其覆盖sum。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 6 struct T{ 7 long long sum, sum1; 8 bool flag; 9 }t[100100 << 2]; 10 int n, m; 11 int c, x, y; 12 long long f[100]; 13 void pushup(int x) 14 { 15 int ls = x << 1, rs = x << 1 | 1; 16 t[x].sum = t[ls].sum + t[rs].sum; 17 t[x].sum1 = t[ls].sum1 + t[rs].sum1; 18 } 19 void pushdown(int x) 20 { 21 if (t[x].flag){ 22 t[x<<1].sum = t[x<<1].sum1; 23 t[x<<1].flag = true; 24 t[x<<1|1].sum = t[x<<1|1].sum1; 25 t[x<<1|1].flag = true; 26 t[x].flag = false; 27 } 28 } 29 void build(int x, int l, int r) 30 { 31 t[x].flag = false; 32 if (l == r){ 33 t[x].flag = t[x].sum = 0; 34 t[x].sum1 = 1; 35 return; 36 } 37 int mid = l + r >> 1; 38 build(x<<1, l, mid); 39 build(x<<1|1, mid+1, r); 40 pushup(x); 41 } 42 void add(int x, int l, int r, int p, long long val) 43 { 44 if (l == r){ 45 t[x].sum += val; 46 if (t[x].sum >= f[91]) 47 t[x].sum1 = f[91]; 48 else if (t[x].sum <= 1) 49 t[x].sum1 = 1; 50 else{ 51 long long * p = upper_bound(f, f+92, t[x].sum); 52 if (t[x].sum - *(p-1) <= *p - t[x].sum) 53 t[x].sum1 = *(p-1); 54 else 55 t[x].sum1 = *p; 56 } 57 return; 58 } 59 int mid = l + r >> 1; 60 pushdown(x); 61 if (p <= mid) 62 add(x<<1, l, mid, p, val); 63 if (p > mid) 64 add(x<<1|1, mid+1, r, p, val); 65 pushup(x); 66 } 67 void tra(int x, int l, int r, int s, int e) 68 { 69 if (l == r){ 70 t[x].sum = t[x].sum1; 71 return; 72 } 73 if (s <= l && r <= e){ 74 t[x].flag = true; 75 t[x].sum = t[x].sum1; 76 return; 77 } 78 int mid = l + r >> 1; 79 pushdown(x); 80 if (s <= mid) 81 tra(x<<1, l, mid, s, e); 82 if (mid < e) 83 tra(x<<1|1, mid+1, r, s, e); 84 pushup(x); 85 } 86 long long query(int x, int l, int r, int s, int e) 87 { 88 if (l == r) 89 return t[x].sum; 90 if (s <= l && r <= e) 91 return t[x].sum; 92 long long tmp = 0; 93 int mid = l + r >> 1; 94 pushdown(x); 95 if (s <= mid) 96 tmp += query(x<<1, l, mid, s, e); 97 if (mid < e) 98 tmp += query(x<<1|1, mid+1, r, s, e); 99 pushup(x); 100 return tmp; 101 } 102 int main() 103 { 104 f[0] = 1; f[1] = 1; 105 for (int i = 2; i <= 91; i++) 106 f[i] = f[i-1] + f[i-2]; 107 while(scanf("%d %d", &n, &m) != EOF) 108 { 109 build(1, 1, n); 110 while(m--){ 111 scanf("%d %d %d", &c, &x, &y); 112 if (c == 1) add(1, 1, n, x, (long long)y); 113 if (c == 2) printf("%I64d\n", query(1, 1, n, x, y)); 114 if (c == 3) tra(1, 1, n, x, y); 115 } 116 } 117 return 0; 118 }
HDOJ 4893 Wow! Such Sequence! 线段树,布布扣,bubuko.com
HDOJ 4893 Wow! Such Sequence! 线段树
原文:http://www.cnblogs.com/james47/p/3876475.html