#include<bits/stdc++.h>
using namespace std;
inline int read()
{
	int x = 0, f = 1;
	char c = getchar();
	while (c < '0' || c > '9')
	{
		if (c == '-') f = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
	{
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x * f;
}
int a[6000000];
int arr[6000000];
int main() {
	int n,q,min1=100000000;
	n=read();
	q=read();
	for(int i=1; i<=n; i++) {
		a[i]=read();
		arr[i]=a[i]-a[i-1];
	}
	for(int i=1; i<=q; i++) {
		int x,y,z;
		x=read();
		y=read();
		z=read();
		arr[x]+=z;
		arr[y+1]-=z;

	}
	for(int i=1; i<=n; i++) {
		arr[i]+=arr[i-1];
		min1=min(arr[i],min1);
	}
	cout<<min1;
	return 0;
}

本题在快读的情况下1,3依旧过不了，选用C++14(O2) 即可。

#include<bits/stdc++.h>
using namespace std;
int d[5000001];
int a[5000001];
inline int read()
{
int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
x = x * 10 + c - '0';
c = getchar();
}
return x*f;
}
int main()
{
int n,p,x,y,z,i,mink;
n=read();p=read();
for(i=1;i<=n;i++)
{
a[i]=read();
}
for(i=1;i<=n;i++)
{
	d[i]=a[i]-a[i-1];
}
for(i=0;i<p;i++)
{
	x=read();
	y=read();
	z=read();
	d[x]+=z;
	d[y+1]-=z;
}
mink=100000;
for(i=1;i<=n;i++)
{
	a[i]=a[i-1]+d[i];
	mink=min(mink,a[i]);
}
cout<<mink;
return 0;
}