题目链接：
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。最后求全部的硬币面值是多少。
直接贴代码啦
code:
#include#include #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; //注意之处。这个区间的值是这么计算的 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