题目链接
P2023 [AHOI2009]维护序列
解题思路
线段树板子。不难，但是...有坑。坑有多深？一页\(WA\)。
由于乘法可能乘\(k=0\)，我这种做法可能会使结果产生负数。于是就有了这篇题解。
（详情见代码注释）
AC代码
#include<stdio.h>
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
typedef long long ll;
int n,m;
ll mod,k,a[500010];
struct Tree{int left,right;ll data,lazy,mul;
}tree[2000010];
void build(int p,int left,int right){tree[p].left=left;tree[p].right=right;tree[p].mul=1;if(left==right){tree[p].data=a[left];return;}build(p<<1,left,(left+right)>>1);build(p<<1|1,((left+right)>>1)+1,right);tree[p].data=(tree[p<<1].data+tree[p<<1|1].data)%mod;
}
void pushdown(int p){ll mul=tree[p].mul,lazy=tree[p].lazy;tree[p<<1].lazy*=mul;tree[p<<1].lazy+=lazy;tree[p<<1].lazy%=mod;tree[p<<1].mul*=mul;tree[p<<1].mul%=mod;tree[p<<1|1].lazy*=mul;tree[p<<1|1].lazy+=lazy;tree[p<<1|1].lazy%=mod;tree[p<<1|1].mul*=mul;tree[p<<1|1].mul%=mod;tree[p].data*=tree[p].mul;tree[p].data+=(tree[p].right-tree[p].left+1)*tree[p].lazy;tree[p].data%=mod;tree[p].lazy=0;tree[p].mul=1;
}
void add(int left,int right,ll k,int p){int l=tree[p].left,r=tree[p].right;if(l>right||r<left||p>4*n)return;pushdown(p);if(l>=left&&r<=right){tree[p].lazy+=k;tree[p].lazy%=mod;return;}tree[p].data+=k*(min(right,r)-max(left,l)+1);tree[p].data%=mod;add(left,right,k,p<<1);add(left,right,k,p<<1|1);
}
ll multy(int left,int right,ll k,int p){int l=tree[p].left,r=tree[p].right;if(l>right||r<left||p>4*n)return 0;pushdown(p);if(l>=left&&r<=right){ll temp=tree[p].data*tree[p].mul+tree[p].lazy*(r-l+1);tree[p].lazy*=k;tree[p].lazy%=mod;tree[p].mul*=k;tree[p].mul%=mod;return ((k-1)*temp%mod+mod)%mod;//非常重要！！！！！！ }ll temp=multy(left,right,k,p<<1)+multy(left,right,k,p<<1|1); tree[p].data+=temp;tree[p].data=;return temp;
}
ll query(int left,int right,int p){int l=tree[p].left,r=tree[p].right;if(l>right||r<left||p>4*n)return 0;pushdown(p);if(l>=left&&r<=right)return tree[p].data;return query(left,right,p<<1)+query(left,right,p<<1|1);
}
int main(){int s,x,y,i;scanf("%d%lld",&n,&mod);for(i=1;i<=n;i++)scanf("%lld",&a[i]);build(1,1,n);scanf("%d",&m);while(m--){scanf("%d%d%d",&s,&x,&y);if(s-3){scanf("%lld",&k);if(s-1)add(x,y,k,1);else multy(x,y,k,1);}else printf("%lld\n",query(x,y,1)%mod);}return 0;
}