题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1698

题目大意：给你三种硬币，分别面值为1,2,3。初始时，面值都为1，后面没更改一次，就把这个区间的银币换位面值为 X的硬币，最后求所有的硬币面值是多少。

直接贴代码啦

code:

#include<stdio.h>
#include<iostream>
#define L(u) (u<<1)
#define R(u) (u<<1|1)

const int M=100010;

struct Node
{
    int l,r;
    int add,sum;
}node[M*4];

//int a[M];

void pushup(int u)
{
    node[u].sum=node[L(u)].sum+node[R(u)].sum;
    return ;
}

void pushdown(int u)
{
    node[L(u)].add=node[u].add;
    node[L(u)].sum=(node[L(u)].r-node[L(u)].l+1)*node[u].add;
    node[R(u)].add=node[u].add;
    node[R(u)].sum=(node[R(u)].r-node[R(u)].l+1)*node[u].add;
    node[u].add=0;
}

void build(int i,int left,int right)
{
    node[i].l=left;
    node[i].r=right;
    node[i].add=0;
    if(node[i].l==node[i].r)
    {
        node[i].sum=1;
        return;
    }
    int mid=(node[i].l+node[i].r)/2;
    build(L(i),left,mid);
    build(R(i),mid+1,right);
    pushup(i);
}

void update(int u,int left,int right,int val)
{
    if(left<=node[u].l&&node[u].r<=right)
    {
        node[u].add=val;                                        
        node[u].sum=(node[u].r-node[u].l+1)*val;<span style="white-space:pre">		</span>//注意之处，这个区间的值是这么计算的<span style="white-space:pre">	</span>
        return ;
    }
    //node[u].sum=(right-left+1)*val;
    if(node[u].add) pushdown(u);
    int mid=(node[u].l+node[u].r)/2;
    if(right<=mid) update(L(u),left,right,val);
    else if(left>mid) update(R(u),left,right,val);
    else
    {
        update(L(u),left,mid,val);
        update(R(u),mid+1,right,val);
    }
    node[u].sum=node[L(u)].sum+node[R(u)].sum;
}

int query(int u,int left,int right)
{
     if(left<=node[u].l&&node[u].r<=right)
    {
        return node[u].sum;
    }
    if(node[u].add) pushdown(u);
    int mid=(node[u].l+node[u].r)/2;
     if(right<=mid) return query(L(u),left,right);
    else if(left>mid) return query(R(u),left,right);
    else return (query(L(u),left,mid)+query(R(u),mid+1,right));
}

int main()
{
    int T,n,m,i,x,y,z;
    int k=0;
    scanf("%d",&T);
    while(T--)
    {
        k++;
         scanf("%d%d",&n,&m);
         build(1,1,n);
         for(i=0;i<m;i++)
         {
             scanf("%d%d%d",&x,&y,&z);
             update(1,x,y,z);
         }
         printf("Case %d: The total value of the hook is %d.\n",k,query(1,1,n));
    }
    return 0;
}