题意：有n个花盆，最开始都是空的，有两种操作，第一种从第i个花盆开始，给你k朵花，然你把空的花盆插上花，插到最后还有花，就把花丢掉，（输出开始插的花盆编号和最后一朵花的花盆编号，如果不能插一朵，就输出Can not put any one.
），第二种操作就是从I到J花盆，清除其中的花（输出清除的花数）。
解题思路：线段树的基本操作，解题思路一是 线段树+ 二分插的位置，还有一种是 直接线段树（利用性质直接查找插花的位置）
解题代码：
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #define MAXN 50005 5 struct node{ 6 int left ,right , mid ; 7 int num , cover ; 8 }tree[MAXN *4]; 9 int L(int c){ 10 return 2 *c ; 11 } 12 int R(int c){ 13 return 2 *c + 1; 14 } 15 void Pushdown(int c){ 16 if(tree[c].cover == 1){ 17 tree[L(c)].cover = 1; 18 tree[L(c)].num = 0; 19 tree[R(c)].cover = 1; 20 tree[R(c)].num = 0; 21 tree[c].cover = 2 ; 22 } 23 else if(tree[c].cover == 0){ 24 tree[L(c)].cover = 0 ; 25 tree[R(c)].cover = 0 ; 26 tree[L(c)].num = (tree[L(c)].right - tree[L(c)].left + 1); 27 tree[R(c)].num = (tree[R(c)].right - tree[R(c)].left + 1); 28 tree[c].cover = 2; 29 }//两种状态，使其跟新后不再跟新 30 } 31 void Pushup(int c) 32 { 33 tree[c].num = tree[L(c)].num + tree[R(c)].num; 34 } 35 void build(int c ,int p , int v ){ 36 tree[c].left = p ; 37 tree[c].right = v ; 38 tree[c].mid = (p+v)/2; 39 tree[c].cover = 0 ; 40 tree[c].num = 0 ; 41 if(p == v){ 42 tree[c].num = 1 ;//空花盆为1，这样好查找 43 return ; 44 } 45 build(L(c),p,tree[c].mid); 46 build(R(c),tree[c].mid+1 ,v); 47 Pushup(c); 48 } 49 int tsum = 0 ; 50 void getsum(int c , int p ,int v ){ 51 if(p <= tree[c].left && v>= tree[c].right){ 52 tsum += tree[c].num ; 53 return ; 54 } 55 Pushdown(c); 56 if(v <= tree[c].mid )getsum(L(c),p,v); 57 else if(p > tree[c].mid) getsum(R(c),p,v); 58 else { 59 getsum(L(c),p,tree[c].mid); 60 getsum(R(c),tree[c].mid + 1,v); 61 } 62 } 63 void update(int c , int p , int v , int value){ 64 // printf("%d %d ",p,v); 65 if (p <= tree[c].left && v >= tree[c].right){ 66 tree[c].num = (tree[c].right - tree[c].left + 1) * (1-value); 67 tree[c].cover = value; 68 return ; 69 } 70 Pushdown(c); 71 if(v <= tree[c].mid ) update(L(c),p , v ,value); 72 else if(p > tree[c].mid) update(R(c),p,v,value); 73 else { 74 update(L(c),p,tree[c].mid,value); 75 update(R(c),tree[c].mid +1 ,v,value); 76 } 77 Pushup(c); 78 } 79 int ok = 0 ; 80 void Find(int c ,int value){ 81 if(tree[c].left == tree[c].right) 82 { 83 ok = tree[c].left; 84 return; 85 } 86 Pushdown(c); 87 if(tree[L(c)].num < value ) 88 Find(R(c),value - tree[L(c)].num); 89 else Find(L(c),value); 90 } 91 92 int main(){ 93 int t ; 94 scanf("%d",&t); 95 while(t--){ 96 int n , m ; 97 scanf("%d %d",&n,&m); 98 build(1,1,n); 99 for(int i = 1;i <= m;i ++){ 100 101 int t1,t2,t3; 102 scanf("%d %d %d",&t1,&t2,&t3); 103 if(t1 == 2){ 104 t2++; 105 t3++; 106 tsum = 0; 107 getsum(1,t2,t3); 108 update(1,t2,t3,0); 109 printf("%d ",(t3-t2+1)-tsum); 110 }else { 111 t2++; 112 tsum = 0 ; 113 int ok1 = 0 ; 114 if(t2 != 1) 115 getsum(1,1,t2-1); 116 if(tsum == tree[1].num) 117 printf("Can not put any one. "); 118 else { 119 Find(1,tsum+1); 120 ok1 = ok ; 121 if(tree[1].num - tsum < t3 ) 122 Find(1,tree[1].num); 123 else Find(1,tsum + t3); 124 update(1,ok1,ok,1); 125 126 printf("%d %d ",ok1-1,ok-1); 127 } 128 129 130 } 131 } 132 133 printf(" "); 134 } 135 return 0 ; 136 }