結果

問題 No.528 10^9と10^9+7と回文
ユーザー 👑 testestest
提出日時 2017-04-02 21:22:49
言語 C
(gcc 13.3.0)
結果
AC  
実行時間 2 ms / 1,000 ms
コード長 879 bytes
コンパイル時間 174 ms
コンパイル使用メモリ 30,976 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-23 18:28:07
合計ジャッジ時間 1,043 ms
ジャッジサーバーID
(参考情報)
judge3 / judge2
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ファイルパターン 結果
other AC * 28
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.c: In function 'main':
main.c:31:9: warning: implicit declaration of function 'scanf' [-Wimplicit-function-declaration]
   31 |         scanf("%s",num);
      |         ^~~~~
main.c:2:1: note: include '<stdio.h>' or provide a declaration of 'scanf'
    1 | #include <assert.h>
  +++ |+#include <stdio.h>
    2 | #define ll long
main.c:31:9: warning: incompatible implicit declaration of built-in function 'scanf' [-Wbuiltin-declaration-mismatch]
   31 |         scanf("%s",num);
      |         ^~~~~
main.c:31:9: note: include '<stdio.h>' or provide a declaration of 'scanf'
main.c:33:14: warning: implicit declaration of function 'strlen' [-Wimplicit-function-declaration]
   33 |         keta=strlen(num);//log10(N)+1
      |              ^~~~~~
main.c:2:1: note: include '<string.h>' or provide a declaration of 'strlen'
    1 | #include <assert.h>
  +++ |+#include <string.h>
    2 | #define ll long
main.c:33:14: warning: incompatible implicit declaration of built-in function 'strlen' [-Wbuiltin-declaration-mismatch]
   33 |         keta=strlen(num);//log10(N)+1
      |              ^~~~~~
main.c:33:14: note: include '<string.h>' or provide a declaration of 'strlen'
main.c:38:9: warning: implicit declaration of function 'printf' [-Wimplicit-function-declaration]
   38 |         printf("%d\n%d",ans1,ans2);
      |         ^~~~~~
main.c:38:9: note: include '<stdio.h>' or provide a declaration of 'printf'
main.c:38:9: warning: incompatible implicit declaration of built-in function 'printf' [-Wbuiltin-declaration-mismatch]
main.c:38:9: note: include '<stdio.h>' or provide a declaration of 'printf'

ソースコード

diff #
プレゼンテーションモードにする

#include <assert.h>
#define ll long
char num[100010];
int keta;
ll pom(ll a,ll n,ll m){return n?pom(a%m*(a%m)%m,n/2,m)*(n&1?a%m:1)%m:1;}
ll f(ll n,ll m){
//numnmod m
int i;
ll s=0;
for(i=0;i<n;i++)s=(s*10+num[i]-'0')%m;
return s;
}
int g(){
//numnum
//1234,1221,11120false4321,1220,11102true
int l=(keta-1)/2,r=keta/2;
while(l>=0){
if(num[l]<num[r])return 0;
if(num[l]>num[r])return 1;
l--;r++;
}
return 0;
}
int main(){
ll MOD1=1e9,MOD2=1e9+7,ans1=0,ans2=0;
int temp;
scanf("%s",num);
assert(num[0]!='0');
keta=strlen(num);//log10(N)+1
assert(1<=keta&&keta<=100001);
temp=g();
ans1=(pom(10,keta/2,MOD1)-1+f(keta-keta/2,MOD1)-temp+MOD1)%MOD1;
ans2=(pom(10,keta/2,MOD2)-1+f(keta-keta/2,MOD2)-temp+MOD2)%MOD2;
printf("%d\n%d",ans1,ans2);
}
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