題目連結

。。。。。模型轉化錯了被自己蠢到哭

就首先這題如果說質因數只有$2$的時候，操作次數肯定是$2$的個數
如果質因數不只有

由於質數除了$2$都是奇數，所以減$1$再質因數分解肯定會多很多

而“能分出的$2$的個數這個東西”是能線性篩出來的

然後這樣就做完了

程式碼：

#include<cstdio>
#include<vector>
#include<queue>
#include<ctime>
#include<algorithm>
#include<cstdlib>
#include<stack>
#include<cstring>
#include<cmath>
using namespace std;

typedef long long LL;

const int INF = 2147483647;
const int maxn = 100010;

LL ans,f[maxn],pri[maxn],tot;
bool mark[maxn];

inline LL getint()
{
    LL ret = 0,f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') ret = ret * 10 + c - '0',c = getchar();
    return ret * f;
}

int main()
{
    #ifdef AMC
        freopen("AMC1.txt","r",stdin);
    #endif
    f[2] = 1;
    for (int i = 2; i <= 100000; i++)
    {
        if (!mark[i])
            pri[++tot] = i , f[i] = i == 2 ? 1 : f[i - 1];
        for (int j = 1; j <= tot; j++)
        {
            if (i * pri[j] > 100000) break;
            mark[i * pri[j]] = 1;
            f[i * pri[j]] = f[i] + f[pri[j]];
            if (i % pri[j] == 0) break;
        }
    }

    int t = getint();
    while (t--)
    {
        ans = 0;
        int n = getint(),flag = 0;
        for (int i = 1; i <= n; i++)
        {
            LL p = getint(),q = getint();
            ans += f[p] * q;
            flag += !(p & 1);
        }
        if (!flag) ans++;
        printf("%lld\n",ans);
    }
    return 0;
}