P1083 借教室

差分数组维护前缀和，之后从后往前判断是否合法，找到尽量位置在后的合法位置，判断输出就好了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<deque>
#include<algorithm>
#define ll long long
using namespace std;
const ll oo=0x3f3f3f3f;
const ll N=1000005;

ll n,m,x=-1,sum;
ll a[N],c[N],l[N],r[N],cf[N];
bool flag=1;

ll Min(ll a,ll b){return a<b?a:b;}
ll Max(ll a,ll b){return a>b?a:b;}
ll Abs(ll a){return a>0?a:-a;}

ll get(){
    char zy=getchar();
    ll z=1,y=0;
    while(zy>'9'||zy<'0'){
        if(zy=='-') z=-1;
        zy=getchar();
    }
    while(zy>='0'&&zy<='9'){
        y=(y<<1)+(y<<3)+zy-'0';
        zy=getchar();
    }
    return z*y;
}

int main(){
    n=get();m=get();
    for(ll i=1;i<=n;i++) a[i]=get();
    for(ll i=1;i<=m;i++){
        c[i]=get();l[i]=get();r[i]=get();
        cf[l[i]]+=c[i];
        cf[r[i]+1]-=c[i];
    }
    ll j=m;
    for(ll i=1;i<=n;i++){
        sum+=cf[i];
        if(sum>a[i]){
            while(sum>a[i]){
                cf[l[j]]-=c[j];
                cf[r[j]+1]+=c[j];
                if(l[j]<=i&&i<=r[j]) sum-=c[j];
                j--;
            }
            if(flag){
                x=j;
                flag=0;
            }
            else x=min(x,j);
        }
    }
    if(x==-1)printf("0
");
    else printf("-1
%lld
",x+1);
    return 0;
}