题意：给你一个数列，给你两个操作，
1）数列中L-R每个值都赋值为 X
2）数列中L-R每个大与 X 的数都变为 gcd（a[i],X） (L <= i <= R)
解题思路：这个题目按理来说时间复杂度是不严谨的，本来想把gcd 从大到小存到 vector 里面，没想到直接更新就行了。
不过从这场比赛发现自己细节方面太糟糕了。还是需要改进
解题代码：
1 // File Name: 1006.cpp 2 // Author: darkdream 3 // Created Time: 2014年07月31日 星期四 12时46分19秒 4 5 #include<vector> 6 #include<list> 7 #include<map> 8 #include<set> 9 #include<deque> 10 #include<stack> 11 #include<bitset> 12 #include<algorithm> 13 #include<functional> 14 #include<numeric> 15 #include<utility> 16 #include<sstream> 17 #include<iostream> 18 #include<iomanip> 19 #include<cstdio> 20 #include<cmath> 21 #include<cstdlib> 22 #include<cstring> 23 #include<ctime> 24 25 using namespace std; 26 #define maxn 100005 27 struct node{ 28 int l , r ; 29 int is1; 30 int limits; 31 }tree[maxn << 2]; 32 int a[maxn]; 33 int L(int x) 34 { 35 return 2*x; 36 } 37 int gcd(int a, int b) 38 { 39 return b == 0 ?a:gcd(b,a%b); 40 } 41 int R(int x) 42 { 43 return 2 * x + 1; 44 } 45 void pushup(int c) 46 { 47 tree[c].limits = max(tree[L(c)].limits,tree[R(c)].limits); 48 } 49 void pushdown(int c) 50 { 51 if(tree[c].is1) 52 { 53 tree[L(c)].is1 = 1; 54 tree[R(c)].is1 = 1; 55 tree[c].is1 = 0; 56 tree[L(c)].limits = tree[R(c)].limits = tree[c].limits; 57 } 58 } 59 void build(int c , int l ,int r) 60 { 61 tree[c].l = l ; 62 tree[c].r = r; 63 tree[c].is1 = 0 ; 64 if(l == r) 65 { 66 tree[c].limits = a[l]; 67 return ; 68 } 69 int m = (l+r)/2; 70 build(L(c),l,m); 71 build(R(c),m+1,r); 72 pushup(c); 73 } 74 void update1(int c, int l , int r,int x) 75 { 76 if(l <= tree[c].l && r >= tree[c].r) 77 { 78 tree[c].is1 = 1; 79 tree[c].limits = x; 80 return ; 81 } 82 int m = (tree[c].l+ tree[c].r)/2; 83 pushdown(c); 84 if(l <= m) 85 update1(L(c),l,r,x); 86 if( r > m ) 87 update1(R(c),l,r,x); 88 pushup(c); 89 } 90 void update2(int c, int l , int r,int x) 91 { 92 if(tree[c].limits < x) 93 { 94 return ; 95 } 96 if(tree[c].is1 && l <= tree[c].l && r >= tree[c].r) 97 { 98 tree[c].limits = gcd(tree[c].limits,x); 99 return ; 100 } 101 if(tree[c].l == tree[c].r) 102 { 103 tree[c].is1 = 0 ; 104 tree[c].limits = gcd(x,tree[c].limits); 105 return; 106 } 107 int m = (tree[c].l+ tree[c].r)/2; 108 pushdown(c); 109 if(l <= m) 110 update2(L(c),l,r,x); 111 if( r > m ) 112 update2(R(c),l,r,x); 113 pushup(c); 114 } 115 void put(int l , int r , int x) 116 { 117 // printf("****%d %d ",l,r); 118 for(int i =l ;i <= r;i ++ ) 119 a[i] = x; 120 return; 121 } 122 void get(int c ) 123 { 124 if(tree[c].is1) 125 { 126 put(tree[c].l,tree[c].r,tree[c].limits); 127 return; 128 } 129 if(tree[c].l == tree[c].r) 130 { 131 a[tree[c].l] = tree[c].limits; 132 return; 133 } 134 pushdown(c); 135 get(L(c)); 136 get(R(c)); 137 } 138 int main(){ 139 int t; 140 //freopen("input.txt","r",stdin); 141 //freopen("output.txt","w",stdout); 142 scanf("%d",&t); 143 while(t--) 144 { 145 //memset(tree,0,sizeof(tree)); 146 int n ,m; 147 scanf("%d",&n); 148 for(int i = 1;i <= n;i ++) 149 { 150 scanf("%d",&a[i]); 151 // printf("%d ",a[i]); 152 } 153 //printf(" "); 154 build(1,1,n); 155 scanf("%d",&m); 156 int ta,tb,tc,td; 157 for(int i = 1;i <= m;i ++) 158 { 159 scanf("%d %d %d %d",&ta,&tb,&tc,&td); 160 if(ta == 1) 161 { 162 update1(1,tb,tc,td); 163 }else{ 164 update2(1,tb,tc,td); 165 } 166 } 167 get(1); 168 for(int i =1;i <= n;i ++) 169 printf("%d ",a[i]); 170 printf(" "); 171 //printf("%d ",t); 172 } 173 return 0; 174 }