first code(求总和):
#include
#include
using namespace std;
#define lchild left, mid, root<<1
#define rchild mid+1, right, root<<1|1
#define maxn 100100
int lazy[maxn<<2];
int sum[maxn<<2];
void update(int root)
{
sum[root] = sum[root<<1]+sum[root<<1|1];
}
void lazyup(int root, int sub)
{
if (lazy[root])
{
lazy[root<<1] = lazy[root<<1|1] = lazy[root];
sum[root<<1] = (sub-sub/2)*lazy[root];
sum[root<<1|1] = (sub/2)*lazy[root];
lazy[root] = 0;
}
}
void build(int left, int right, int root)
{
lazy[root] = 0;
sum[root] = 1;
if (left==right)
return;
int mid = (left+right)>>1;
build(lchild);
build(rchild);
update(root);
}
void operate(int l, int r, int c, int left, int right, int root)
{
if (l<=left&&r>=right)
{
lazy[root] = c;
sum[root] = c*(right-left+1);
return;
}
lazyup(root, right-left+1);
int mid = (left+right)>>1;
if (l<=mid)
operate(l,r,c,lchild);
if (r>mid)
operate(l,r,c,rchild);
update(root);
}
int main()
{
int N,n,m;
cin>>N;
int t=N;
while (N--)
{
scanf("%d%d",&n,&m);
build (1,n,1);
while (m--)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
operate(a,b,c,1,n,1);
}
printf("Case %d: The total value of the hook is %d.\n", t-N, sum[1]);
}
return 0;
}
second code(求每个) :
#include
#include
using namespace std;
int ans[1000000],n;
struct node
{
int l,r,n;
}a[1000000];
void build(int left, int right ,int root)
{
a[root].l = left;
a[root].r = right;
a[root].n = 0;
if (left!=right)
{
int mid = (left+right)>>1;
build(left, mid, root<<1);
build(mid+1, right, root<<1|1);
}
}
void operate(int left, int right, int root)
{
if (a[root].l==left&&a[root].r==right)
{
a[root].n++;
}
else
{
int mid = (a[root].l+a[root].r)>>1;
if (right<=mid)
{
operate(left,right,root<<1);
}
else if(left>mid)
{
operate(left,right,root<<1|1);
}
else
{
operate(left,mid,root<<1);
operate(mid+1,right,root<<1|1);
}
}
}
void add(int root)
{
int i;
for(i = a[root].l; i<=a[root].r; i++)//该区间所有编号都被刷了一次
ans[i]+=a[root].n;
if(a[root].l == a[root].r)
return;
add(root<<1);
add(root<<1|1);
}
int main()
{
int n,i,j;
while (scanf("%d",&n),n!=0)
{
build(1,n,1);
int x,y;
for(int i = 1; i<=n; i++)
{
scanf("%d%d",&x,&y);
operate(x,y,1);
}
memset(ans,0,sizeof(ans));
add(1);
printf("%d",ans[1]);
for(i = 2; i<=n; i++)
printf(" %d",ans[i]);
printf("\n");
}
return 0;
}