【思维/构造】D

D - Decrease the Sum of Digits

只要发现进位可以直接抹掉后面的位数就行了，剩下的就是在第几位进位的问题。写得繁琐了一点。

int main()
{
    //ios::sync_with_stdio(false);
    //while (scanf("%d%d",&n,&m)!=EOF){
    int t; cin >> t; while (t--) {
        LL n; int s;
        cin >> n >> s;
        string str = to_string(n);
        int len = str.length();
        int cnt = 0;
        for (int i = 0; i < len; i++) {
            cnt += str[i] - '0';
        }
        if (cnt <= s) {
            cout << 0 << endl;
            continue;
        }
        cnt = 0;
        //第一位数>=s，必须进位
        if (str[0] - '0' >= s) {
            //从后往前找到第一个需要借位做减法的位置
            int mark = -1;
            for (int k = len - 1; k > 0; k--) {
                if (str[k] - '0' != 0) {
                    mark = k;
                    break;
                }
            }
            //不需要借位做减法
            if (mark == -1) {
                cout << 10 - (str[0] - '0');
                for (int k = 1; k < len; k++) cout << 0;
                cout << endl;
                continue;
            }
            else {
                //记录当前位是否会减出前导零
                int first = 1;
                for (int k = 0; k < len; k++) {
                    if (k < mark) {
                        //9-str[k]会减出前导零，跳过
                        if (first && str[k] == '9') continue;
                        else {
                            cout << 9 - (str[k] - '0');
                            first = 0;
                        }
                    }
                    else if (k == mark) cout << 10 - (str[k] - '0');
                    else if (k > mark) cout << 0;
                }
                cout << endl;
                continue;
            }
        }
        else {
            for (int i = 0; i < len; i++) {
                cnt += str[i] - '0';
                //累加到第i位之后超出s，在第i位往前进一位
                //如235 5，加到第1位（从0起计）后=5，则应进位成300，答案即300-235
                if (cnt >= s) {
                    int mark = -1;
                    for (int k = len - 1; k > i; k--) {
                        if (str[k] != '0') {
                            mark = k;
                            break;
                        }
                    }
                    if (mark == -1) {
                        cout << 10 - (str[i] - '0');
                        for (int k = i + 1; k < len; k++) cout << 0;
                        cout << endl;
                        break;
                    }
                    else {
                        int first = 1;
                        for (int k = i; k < len; k++) {
                            if (k < mark) {
                                if (first && str[k] == '9') continue;
                                else {
                                    cout << 9 - (str[k] - '0');
                                    first = 0;
                                }
                            }
                            else if (k == mark) cout << 10 - (str[k] - '0');
                            else if (k > mark) cout << 0;
                        }
                        cout << endl;
                        break;
                    }
                }
            }
        }
    }
    return 0;
}