### AC代码：

`#include <cstdio>#include <vector>#include <queue>#include <cstring>#include <cmath>#include <map>#include <set>#include <string>#include <iostream>#include <algorithm>#include <iomanip>#include <stack>#include <queue>using namespace std;#define sd(n) scanf("%d", &n)#define sdd(n, m) scanf("%d%d", &n, &m)#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)#define pd(n) printf("%d\n", n)#define pc(n) printf("%c", n)#define pdd(n, m) printf("%d %d\n", n, m)#define pld(n) printf("%lld\n", n)#define pldd(n, m) printf("%lld %lld\n", n, m)#define sld(n) scanf("%lld", &n)#define sldd(n, m) scanf("%lld%lld", &n, &m)#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)#define sf(n) scanf("%lf", &n)#define sc(n) scanf("%c", &n)#define sff(n, m) scanf("%lf%lf", &n, &m)#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)#define ss(str) scanf("%s", str)#define rep(i, a, n) for (int i = a; i <= n; i++)#define per(i, a, n) for (int i = n; i >= a; i--)#define mem(a, n) memset(a, n, sizeof(a))#define debug(x) cout << #x << ": " << x << endl#define pb push_back#define all(x) (x).begin(), (x).end()#define fi first#define se second#define mod(x) ((x) % MOD)#define gcd(a, b) __gcd(a, b)#define lowbit(x) (x & -x)typedef pair<int, int> PII;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int MOD = 1e9 + 7;const double eps = 1e-9;const ll INF = 0x3f3f3f3f3f3f3f3fll;const int inf = 0x3f3f3f3f;inline int read(){    int ret = 0, sgn = 1;    char ch = getchar();    while (ch < '0' || ch > '9')    {        if (ch == '-')            sgn = -1;        ch = getchar();    }    while (ch >= '0' && ch <= '9')    {        ret = ret * 10 + ch - '0';        ch = getchar();    }    return ret * sgn;}inline void Out(int a) //Êä³öÍâ¹Ò{    if (a > 9)        Out(a / 10);    putchar(a % 10 + '0');}ll gcd(ll a, ll b){    return b == 0 ? a : gcd(b, a % b);}ll lcm(ll a, ll b){    return a * b / gcd(a, b);}///快速幂m^k%modll qpow(ll a, ll b, ll mod){    if (a >= mod)        a = a % mod + mod;    ll ans = 1;    while (b)    {        if (b & 1)        {            ans = ans * a;            if (ans >= mod)                ans = ans % mod + mod;        }        a *= a;        if (a >= mod)            a = a % mod + mod;        b >>= 1;    }    return ans;}// 快速幂求逆元int Fermat(int a, int p) //费马求a关于b的逆元{    return qpow(a, p - 2, p);}///扩展欧几里得ll exgcd(ll a, ll b, ll &x, ll &y){    if (b == 0)    {        x = 1;        y = 0;        return a;    }    ll g = exgcd(b, a % b, x, y);    ll t = x;    x = y;    y = t - a / b * y;    return g;}char s[10000];ll ans = 0;int main(){    while (~ss(s + 1))    {        if (s[1] == '|')            ans += 1ll * 42 * strlen(s + 1);        else        {            ll res = 0;            int pos = 1;            while (isdigit(s[pos]))            {                res *= 10;                res += s[pos++] - '0';            }            ll len = strlen(s + 1) - pos - 1;            ans += res * max(len, 1ll);        }    }    if (ans % 10 != 0 && ans % 10 > 4)    {        ans -= ans % 10;        ans += 10;    }    else if (ans % 10 != 0 && ans % 10 < 5)    {        ans -= ans % 10;    }    printf("%lld,-\n", ans);}`