hdu 4578 Transformation
http://acm.hdu.edu.cn/showproblem.php?pid=4578
题意：1,a,b,c代表在a,b区间的每一个数加上c；2,a,b,c代表在a,b区间的每一个数乘上c； 3,a,b,c代表在a,b区间的每一个数变为c；4,a,b,c是求在a,b区间的每一个数的c次方的和。
先乘后加。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define maxn 100010 5 using namespace std; 6 7 int n,m; 8 struct node 9 { 10 int r,l; 11 int sum; 12 int add; 13 int mul; 14 int sum2; 15 int sum3; 16 }tree[maxn*6]; 17 18 void build(int i,int l,int r) 19 { 20 tree[i].l=l; 21 tree[i].r=r; 22 tree[i].sum=0; 23 tree[i].add=0; 24 tree[i].mul=1; 25 tree[i].sum2=0; 26 tree[i].sum3=0; 27 if(l==r) return ; 28 int mid=(l+r)>>1; 29 build(i<<1,l,mid); 30 build(i<<1|1,mid+1,r); 31 } 32 33 void mull(int i,int data1,int data) 34 { 35 (tree[i].sum*=data1)%=10007; 36 tree[i].sum2=tree[i].sum2*data1%10007*data1%10007; 37 tree[i].sum3=tree[i].sum3*data1%10007*data1%10007*data1%10007; 38 (tree[i].mul*=data1)%=10007; 39 (tree[i].add*=data1)%=10007; 40 (tree[i].sum3+=((tree[i].r-tree[i].l+1)%10007*(data%10007*data%10007*data%10007)))%=10007; 41 (tree[i].sum3+=3*data%10007*tree[i].sum2%10007)%=10007; 42 (tree[i].sum3+=(3*data%10007*data%10007*tree[i].sum%10007))%=10007; 43 (tree[i].sum2+=(tree[i].r-tree[i].l+1)*(data*data%10007)%10007)%=10007; 44 (tree[i].sum2+=(2*data%10007*tree[i].sum%10007))%=10007; 45 (tree[i].sum+=(tree[i].r-tree[i].l+1)%10007*data%10007)%=10007; 46 (tree[i].add+=data)%=10007; 47 } 48 void down(int i) 49 { 50 if(tree[i].l==tree[i].r) return ; 51 mull(i<<1,tree[i].mul,tree[i].add); 52 mull(i<<1|1,tree[i].mul,tree[i].add); 53 tree[i].add=0; 54 tree[i].mul=1; 55 } 56 void update(int i,int l,int r,int data1,int data) 57 { 58 if(tree[i].l==l&&tree[i].r==r) 59 { 60 mull(i,data1,data); 61 return ; 62 } 63 down(i); 64 int mid=(tree[i].l+tree[i].r)>>1; 65 if(r<=mid) 66 { 67 update(i<<1,l,r,data1,data); 68 } 69 else if(l>mid) 70 { 71 update(i<<1|1,l,r,data1,data); 72 } 73 else 74 { 75 update(i<<1,l,mid,data1,data); 76 update(i<<1|1,mid+1,r,data1,data); 77 } 78 tree[i].sum=(tree[i<<1].sum+tree[i<<1|1].sum)%10007; 79 tree[i].sum2=(tree[i<<1].sum2+tree[i<<1|1].sum2)%10007; 80 tree[i].sum3=(tree[i<<1].sum3+tree[i<<1|1].sum3)%10007; 81 } 82 83 int search1(int i,int l,int r,int ch) 84 { 85 if(tree[i].l==l&&tree[i].r==r) 86 { 87 if(ch==1) 88 return tree[i].sum; 89 else if(ch==2) 90 return tree[i].sum2; 91 else if(ch==3) 92 return tree[i].sum3; 93 } 94 down(i); 95 int mid=(tree[i].l+tree[i].r)>>1; 96 if(r<=mid) 97 { 98 return search1(i<<1,l,r,ch); 99 } 100 else if(l>mid) 101 { 102 return search1(i<<1|1,l,r,ch); 103 } 104 else 105 { 106 return (search1(i<<1,l,mid,ch)+search1(i<<1|1,mid+1,r,ch))%10007; 107 } 108 } 109 110 int main() 111 { 112 while(scanf("%d%d",&n,&m)!=EOF) 113 { 114 if(n==0&&m==0) break; 115 build(1,1,n); 116 for(int i=1; i<=m; i++) 117 { 118 int x,y,op,c; 119 scanf("%d%d%d%d",&op,&x,&y,&c); 120 if(op==1) 121 { 122 update(1,x,y,1,c); 123 } 124 else if(op==2) 125 { 126 update(1,x,y,c,0); 127 } 128 else if(op==3) 129 { 130 update(1,x,y,0,c); 131 } 132 else if(op==4) 133 { 134 printf("%d\n",search1(1,x,y,c)%10007); 135 } 136 } 137 } 138 return 0; 139 }
