H - easy - Roundgod and Milk Tea

题目

Roundgod is a famous milk tea lover at Nanjing University second to none. This year, he plans to conduct a milk tea festival. There will be $n$ classes participating in this festival, where the $i$th class has $a_i$ students and will make $b_i$ cups of milk tea.

Roundgod wants more students to savor milk tea, so he stipulates that every student can taste at most one cup of milk tea. Moreover, a student can’t drink a cup of milk tea made by his class. The problem is, what is the maximum number of students who can drink milk tea?

Input

The first line of input consists of a single integer $T$ $(1 \leq T \leq 25)$, denoting the number of test cases.

Each test case starts with a line of a single integer $n$ $(1 \leq n \leq 10^6)$, the number of classes. For the next $n$ lines, each containing two integers $a, b$ $(0 \leq a, b \leq 10^9)$, denoting the number of students of the class and the number of cups of milk tea made by this class, respectively.

It is guaranteed that the sum of $n$ over all test cases does not exceed $6 \times 10^6$.

Output

For each test case, print the answer as a single integer in one line.

Sample Input

1
2
3 4
2 1

Sample Output

3
~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ 分 ~ 割 ~ 线 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
本来这道题是让队友做的，然后自己在做前面的C题。然而没做出来
结束后开始补题
思路吗。。。。。emmm。。。。
基本没有。。。
一开始打完输入之后，就没什么思路了。
后来，发现每个班级的最大人数是min(a[i],sumb-b[i])。当然是听老师讲过的
就开始打主要模块

		for (i=1;i<=n;i++)
		{
			l=min(m-b[i],a[i]);
			p+=l;
			m-=l;
		}//习惯不好 m是sumb p是ans l是当前班级最大人数

打完测试，发现不对。貌似没把第i个班之前被喝的奶茶减掉。
开始改
改完如下

		for (i=1;i<=n;i++)
		{
			k=max(b[i]-p,(ll)0);
			l=min(m-k,a[i]);
			p+=l;
			m-=l;
		}

然后就AC啦(ﾋ•ω•ﾏ)/
完整代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll T,n,m,i,j,k,l,o,p,a[1000010],b[1000010];
int main()
{
	scanf("%lld",&T);
	while(T--)
	{
		scanf("%lld",&n);
		m=0;p=0;
		for (i=1;i<=n;i++)
		 scanf("%lld%lld",a+i,b+i),m+=b[i];
		for (i=1;i<=n;i++)
		{
			k=max(b[i]-p,(ll)0);
			l=min(m-k,a[i]);
			p+=l;
			m-=l;
		}
		printf("%lld\n",p);
	}
}

把作业水完了