Codeforces 1028E(模拟)

Codeforces 1028E
题意:给你一个环$b$,要你构造一个环$a$,满足$a_i \% a_{i + 1}=b_i$

设$max(b)=mks$,要找到一个数$i$使得$b_i=mks$并且$b_{i-1} < b_i$
然后让$a_i=b_i$,然后后面的就按照$a_i=b_{i+1}+b_i$来算。对于前面的约束也很清楚了,否则余数不对。
然后就是注意无解当且仅当所有数相同并且不是$0$。
然后还要对$0$作出特判。
1、所有数都是$0$,直接输出$n$个相同数即可
2、如果当前$b_i=0​$并且$a_i=mks​$,那么$a_i​$要乘$2​$,否则无法通过下面数据

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0 0 0 100 0

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#include<cstdio>
#include<algorithm>
#include<cstring>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define ULL unsigned long long
#define db double
LL n, b[200000 + 5], a[200000 + 5], pos = 1, flag = true, mks;
void clean() {
}
int solve() {
clean();
for (LL i = 1; i <= n; i++) {
scanf("%lld", &b[i]);
if (b[pos] < b[i]) pos = i;
if (i != 1 && b[i] != b[i - 1]) flag = false;
}
if (flag && b[1] > 0) return printf("NO\n"), 0;//
mks = b[pos];
for (LL i = 1; i <= n; i++) {
LL now;
if (i - 1 <= 0) now = n; else now = i - 1;
if (b[i] != b[now] && b[i] == mks) {
pos = i; break;
}
}
a[pos] = b[pos];
if (a[pos] == 0) {
printf("YES\n");
for (LL i = 1; i <= n; i++) printf("%lld ", 233ll);
return 0;
}//0
LL cnt = 1;
while (cnt <= n - 1) {
LL now;
if (pos - 1 <= 0) now = n; else now = pos - 1;
a[now] = a[pos] + b[now];
if (b[now] == 0 && a[now] == mks) a[now] *= 2ll;//0
cnt++, pos = now;
}
printf("YES\n");
for (LL i = 1; i <= n; i++) printf("%lld ", a[i]);
return 0;
}
int main() {
scanf("%lld", &n), solve();
return 0;
}
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