标签:scanf stream with imp else list tor ret oid
思路:线段树,区间更新,区间查找
1 #include<iostream> 2 #include<vector> 3 #include<string> 4 #include<cmath> 5 #include<set> 6 #include<algorithm> 7 #include<cstdio> 8 #include<map> 9 #include<cstring> 10 #include<list> 11 12 #define MAXSIZE 100010 13 14 using namespace std; 15 16 long long tree[4*MAXSIZE]; 17 long long lz[4*MAXSIZE]; 18 int N, Q; 19 20 void init() 21 { 22 memset(tree, 0, sizeof(tree)); 23 memset(lz, 0, sizeof(lz)); 24 } 25 26 void build(int node, int l, int r) 27 { 28 if(l == r) 29 { 30 scanf("%lld", &tree[node]); 31 return; 32 } 33 34 int mid = (l+r)/2; 35 build(node*2, l, mid); 36 build(node*2+1, mid+1, r); 37 38 tree[node] = tree[node*2] + tree[node*2+1]; 39 } 40 41 void push_down(int node, int l, int r) 42 { 43 if(lz[node]) 44 { 45 int mid = (l+r)/2; 46 lz[node*2] += lz[node]; 47 lz[node*2+1] += lz[node]; 48 tree[node*2] += (mid-l+1)*lz[node]; 49 tree[node*2+1] += (r-mid)*lz[node]; 50 lz[node] = 0; 51 } 52 } 53 54 void update_range(int node, int l, int r, int L, int R, int add) 55 { 56 if(l <= L && r >= R) 57 { 58 lz[node] += add; 59 tree[node] += (R-L+1)*add; 60 return; 61 } 62 push_down(node, L, R); 63 int mid = (L+R)/2; 64 if(mid >= l) 65 update_range(node*2, l, r, L, mid, add); 66 if(mid < r) 67 update_range(node*2+1, l, r, mid+1, R, add); 68 tree[node] = tree[node*2] + tree[node*2+1]; 69 } 70 71 long long query_range(int node, int l, int r, int L, int R) 72 { 73 if(l <= L && r >= R) 74 { 75 return tree[node]; 76 } 77 push_down(node, L, R); 78 int mid = (L+R)/2; 79 long long sum = 0; 80 if(mid >= l) 81 sum += query_range(node*2, l, r, L, mid); 82 if(mid < r) 83 sum += query_range(node*2+1, l, r, mid+1, R); 84 85 return sum; 86 } 87 88 89 90 int main() 91 { 92 scanf("%d%d", &N, &Q); 93 init(); 94 build(1, 1, N); 95 while(Q--) 96 { 97 char ch; 98 scanf(" %c", &ch); 99 if(ch == ‘Q‘) 100 { 101 int a, b; 102 scanf("%d%d", &a, &b); 103 printf("%lld\n", query_range(1, a, b, 1, N)); 104 } 105 else if(ch == ‘C‘) 106 { 107 int a, b, c; 108 scanf("%d%d%d", &a, &b, &c); 109 update_range(1, a, b, 1, N, c); 110 } 111 } 112 113 114 return 0; 115 }
A Simple Problem with Integers POJ - 3468 (线段树)
标签:scanf stream with imp else list tor ret oid
原文地址:https://www.cnblogs.com/FengZeng666/p/11458279.html