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 }

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