题目链接：ZOJ 2706 Thermal Death of the Universe (线段树)

题意：n个数。m个操作。

每一个操作（a,b）表示（a,b）全部值更新为这个区间的平均数：1.当前的数列总和小于等于原数列总和。取平均值的上界，反之。取下界。

注意有负数的情况。

AC代码：

#include<stdio.h>
#include <math.h>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=30010;
const LL INF=0xffffffffffff;
struct node
{LL l,r,laz;LL sum;LL mid(){return (l+r)/2;}
};
struct node tree[maxn<<2];
LL pre;
void build(LL l,LL r,LL rt)
{tree[rt].l=l;tree[rt].r=r;tree[rt].laz=INF;if(tree[rt].l==tree[rt].r){scanf("%lld",&tree[rt].sum);return ;}LL m=tree[rt].mid();build(lson);build(rson);tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
}void PushDown(LL rt,LL m)
{if(tree[rt].laz!=INF){tree[rt<<1].laz=tree[rt].laz;tree[rt<<1|1].laz=tree[rt].laz;tree[rt<<1].sum=(m-(m>>1))*tree[rt].laz;tree[rt<<1|1].sum=(m>>1)*tree[rt].laz;tree[rt].laz=INF;}
}void updata(LL L,LL R,LL add,LL rt)
{if(L<=tree[rt].l && tree[rt].r<=R){tree[rt].laz=add;tree[rt].sum=(tree[rt].r-tree[rt].l+1)*add;return ;}LL m=tree[rt].mid();PushDown(rt,tree[rt].r-tree[rt].l+1);if(R<=m)updata(L,R,add,rt<<1);else if(L>m)updata(L,R,add,rt<<1|1);else{updata(L,R,add,rt<<1);updata(L,R,add,rt<<1|1);}tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
}LL query(LL L,LL R,LL rt)
{if(L<=tree[rt].l && tree[rt].r<=R){return tree[rt].sum;}LL m=tree[rt].mid();PushDown(rt,tree[rt].r-tree[rt].l+1);LL ret=0;if(R<=m)ret+=query(L,R,rt<<1);else if(L>m)ret+=query(L,R,rt<<1|1);else{ret+=query(L,R,rt<<1);ret+=query(L,R,rt<<1|1);}return ret;
}int main()
{LL a,b,i;LL n,m;LL ans;double ave;//printf("%lld\n",INF);while(scanf("%lld %lld",&n,&m)!=EOF){//memset(tree,0,sizeof tree);build(1,n,1);pre=0;for(i=1;i<=n;i++)pre+=query(i,i,1);while(m--){scanf("%lld%lld",&a,&b);LL sum=query(a,b,1);ave=1.0*sum/(b-a+1);if(query(1,n,1)<=pre)ans=(LL)ceil(ave);elseans=(LL)floor(ave);updata(a,b,ans,1);}for(i=1;i<n;i++)printf("%lld ",query(i,i,1));printf("%lld\n\n",query(i,i,1));}
return 0;
}
/*
6 1
1 2 3 4 5 6
1 6
6 2
1 2 3 4 5 6
2 6
1 5
1 11 1
1
1 16 2
1 1 1 3 1 1
3 4
4 56 4
1 2 3 4 5 6
1 2
2 6
1 3
2 53 3
-1 -1 -2
1 2
2 3
1 3
*/