ACM ICPC Vietnam National Second Round D X = X + X % 100

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//ACM ICPC Vietnam National Second Round D////题目大意:////X = X + X % 100.给你一个N和K.求K次操作后N的值////解题思路:////看到N和K都是1e9这种数.想到肯定会有循环节.然后//打表找规律.一开始发现是20一循环.即4的倍数才可以开//始循环,然而太年轻,交了无数发,wrong了无数发.后来发现//10这类的也是有循环4一循环.欣喜的敲了一发,然后还是wrong//这时候,我又想到了00这种特殊的情况,好,又交,又wrong//当然50也要特判啊.25啊,75啊都特判.写完之后,我发现自己太//年轻了,这样就可以过了.还是自己考虑问题不够周全吧.//情况多多,做练习赛的时候并没有做出来.不得不说自己太弱了.//继续加油,FIGHTING!!!#include <cstring>#include <algorithm>#include <iostream>#include <cstdio>#include <cmath>#include <string>#include <vector>#include <queue>#define For(x,a,b,c) for (int x = a; x <= b; x += c)#define Ffor(x,a,b,c) for (int x = a; x >= b; x -= c)#define cls(x,a) memset(x,a,sizeof(x))using namespace std;typedef long long ll;const int MAX_N = 108;const int INF = 0x3f3f3f3f;const ll MOD = 1e6;ll N,K;int a[108];int cnt[108];bool vis[108];int main(){  //  cls(a,0); //打表所用  //  cls(cnt,0);  //  for (int i = 1 ;i < 100;i ++){  //      cls(vis,0);  //      int k = 0;  //      vis[i] = 1;  //      int x = i;  //      x = x + x % 100;  //      int t = x % 100;  //      int c = 1;  //      while(!vis[t]){  //          vis[t] = 1;  //          x = x + x % 100;  //          t = x % 100;  //          c++;  //      }  //      a[i] = x;  //      cnt[i] = c;  //  //  if (i % 10 == 5)  //  //      printf("%d %d %d\n",i,x,c);  //  }    //freopen("1.in","r",stdin);      int kase;    scanf("%d",&kase);    for (int i = 1;i <= kase ;i ++){        scanf("%I64d%I64d",&N,&K);                ll ans = N / 100 * 100;        ll res = N % 100;        if (res == 0){            printf("%I64d\n",N);            continue;        }        if ( res == 50){            printf("%I64d\n", N + 50);            continue;        }        if ( res == 25){            if (K >= 1000000){                printf("%I64d\n",ans + 100);            }            continue;        }        if (res == 75){            if (K >= 1000000){                printf("%I64d\n",ans + 200);            }            continue;        }            if (K < 1000000){            for (int j = 1;j <= K;j ++){                res = res + res % 100;            }            printf("%I64d\n",ans + res);            continue;        }        int c = 0;        if ( res % 10 != 0 || res % 10 != 5){            while((res % 100) % 4){                res += res % 100;                c++;            }            K -= c;            res = res + K / 20 * 1000;            K %= 20;        }        else {            while(((res % 100) / 10) % 2){                res += res % 100;                c++;            }            K -= c;            res = res + K / 4 * 200;            K %= 4;        }        for (int j = 0 ; j < K; j++){            res += res % 100;        }        printf("%I64d\n",ans + res);    }}
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